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Is it nonsense to say B -> [A -> B]?


Suspending some of the usual laws of logicWhat are the dialetheist semantics for logical negation?Is something creating itself nonsense?How do disjunctive antecedents work in Marc Lange's stability concept of laws of nature?Implication Introduction formulated as a theorem?Is there something wrong in breaking the symmetry of Natural Deduction?Proof for the Rule of Absorption in Propositional Logic?Why do contradictions imply anything?Making 'sense' of Wittgenstein's senselessness / nonsense distinction in the TractatusWhat CAN a sentence say about itself? Can a sentence say about itself that it is false?













3















Is it nonsense to say B → [A → B]? I would have thought so, it seems to say that every fact deductively follows from every other. But I was looking at a recent closed question, on natural deduction, and my thinking about it (not as natural deduction) got be wondering.



It seems to me that B with the law of excluded middle and disjunction introduction and conjunction introduction ⇒ [[~A ∧ B] v [A ∧ B]] which with law of non contradiction and disjunctive syllogism ⇒ [A → [A ∧ B]] which with conjunction elimination ⇒ [A → B].



B → [A → B]



Sorry for being so naive, but, assuming I can be understood here, what have I worked out wrong?










share|improve this question



















  • 5





    It doesn't say that "every fact deductively follows from every other", it says that if you have already deduced B, you can deduce it from anything else.

    – Eliran
    9 hours ago











  • is there a phrase for that. one that i can google @Eliran ?

    – another_name
    9 hours ago











  • I do not think this tautology has an established name. If you rewrite it as B→(¬A∨B) it is just an instance of disjunction introduction. Alternatively, if B is derivable from nothing then it surely is derivable from some A, whatever A is. In particular, it is even intuitionistically valid.

    – Conifold
    5 hours ago















3















Is it nonsense to say B → [A → B]? I would have thought so, it seems to say that every fact deductively follows from every other. But I was looking at a recent closed question, on natural deduction, and my thinking about it (not as natural deduction) got be wondering.



It seems to me that B with the law of excluded middle and disjunction introduction and conjunction introduction ⇒ [[~A ∧ B] v [A ∧ B]] which with law of non contradiction and disjunctive syllogism ⇒ [A → [A ∧ B]] which with conjunction elimination ⇒ [A → B].



B → [A → B]



Sorry for being so naive, but, assuming I can be understood here, what have I worked out wrong?










share|improve this question



















  • 5





    It doesn't say that "every fact deductively follows from every other", it says that if you have already deduced B, you can deduce it from anything else.

    – Eliran
    9 hours ago











  • is there a phrase for that. one that i can google @Eliran ?

    – another_name
    9 hours ago











  • I do not think this tautology has an established name. If you rewrite it as B→(¬A∨B) it is just an instance of disjunction introduction. Alternatively, if B is derivable from nothing then it surely is derivable from some A, whatever A is. In particular, it is even intuitionistically valid.

    – Conifold
    5 hours ago













3












3








3








Is it nonsense to say B → [A → B]? I would have thought so, it seems to say that every fact deductively follows from every other. But I was looking at a recent closed question, on natural deduction, and my thinking about it (not as natural deduction) got be wondering.



It seems to me that B with the law of excluded middle and disjunction introduction and conjunction introduction ⇒ [[~A ∧ B] v [A ∧ B]] which with law of non contradiction and disjunctive syllogism ⇒ [A → [A ∧ B]] which with conjunction elimination ⇒ [A → B].



B → [A → B]



Sorry for being so naive, but, assuming I can be understood here, what have I worked out wrong?










share|improve this question
















Is it nonsense to say B → [A → B]? I would have thought so, it seems to say that every fact deductively follows from every other. But I was looking at a recent closed question, on natural deduction, and my thinking about it (not as natural deduction) got be wondering.



It seems to me that B with the law of excluded middle and disjunction introduction and conjunction introduction ⇒ [[~A ∧ B] v [A ∧ B]] which with law of non contradiction and disjunctive syllogism ⇒ [A → [A ∧ B]] which with conjunction elimination ⇒ [A → B].



B → [A → B]



Sorry for being so naive, but, assuming I can be understood here, what have I worked out wrong?







logic






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 hours ago









Joshua

48829




48829










asked 10 hours ago









another_nameanother_name

261114




261114







  • 5





    It doesn't say that "every fact deductively follows from every other", it says that if you have already deduced B, you can deduce it from anything else.

    – Eliran
    9 hours ago











  • is there a phrase for that. one that i can google @Eliran ?

    – another_name
    9 hours ago











  • I do not think this tautology has an established name. If you rewrite it as B→(¬A∨B) it is just an instance of disjunction introduction. Alternatively, if B is derivable from nothing then it surely is derivable from some A, whatever A is. In particular, it is even intuitionistically valid.

    – Conifold
    5 hours ago












  • 5





    It doesn't say that "every fact deductively follows from every other", it says that if you have already deduced B, you can deduce it from anything else.

    – Eliran
    9 hours ago











  • is there a phrase for that. one that i can google @Eliran ?

    – another_name
    9 hours ago











  • I do not think this tautology has an established name. If you rewrite it as B→(¬A∨B) it is just an instance of disjunction introduction. Alternatively, if B is derivable from nothing then it surely is derivable from some A, whatever A is. In particular, it is even intuitionistically valid.

    – Conifold
    5 hours ago







5




5





It doesn't say that "every fact deductively follows from every other", it says that if you have already deduced B, you can deduce it from anything else.

– Eliran
9 hours ago





It doesn't say that "every fact deductively follows from every other", it says that if you have already deduced B, you can deduce it from anything else.

– Eliran
9 hours ago













is there a phrase for that. one that i can google @Eliran ?

– another_name
9 hours ago





is there a phrase for that. one that i can google @Eliran ?

– another_name
9 hours ago













I do not think this tautology has an established name. If you rewrite it as B→(¬A∨B) it is just an instance of disjunction introduction. Alternatively, if B is derivable from nothing then it surely is derivable from some A, whatever A is. In particular, it is even intuitionistically valid.

– Conifold
5 hours ago





I do not think this tautology has an established name. If you rewrite it as B→(¬A∨B) it is just an instance of disjunction introduction. Alternatively, if B is derivable from nothing then it surely is derivable from some A, whatever A is. In particular, it is even intuitionistically valid.

– Conifold
5 hours ago










4 Answers
4






active

oldest

votes


















4














When proving a conditional one assumes the antecedent, B. The goal is not to derive this, but from this assumption to derive the consequent which happens to be A > B.



But A > B is another conditional. Since it is a conditional one derives that in the same way. First assume the antecedent, A. Can one derive B? Yes, one can, because in this derivation we already assumed B in the first line. We can "reiterate" it in the next line as B.



Since from A one can derive B in a subproof, one can introduce a conditional (→I), A > B. This is the same as saying: if I have A, I can derive B. It is a shorthand for that subproof.



But if one has A > B one can introduce the conditional again and write B > [A > B].



Here's the result using the proof checker linked to below:



enter image description here



What this shows is that the argument is valid. The argument might be nonsense depending on what A and B are. Suppose A is Unicorns are white and B is The sky is blue. What this says is If the sky is blue then if unicorns are white then the sky is blue. Suppose the sky is blue. It doesn't matter whether unicorns are white or not. The sky is still blue.



For the reiteration rule (R) see section 16.1 of the forallx text below.




Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/






share|improve this answer























  • thanks, i think i followed the answer, though it's not really in terms i feel comfortable with. you did lose me a bit when talking about unicorns. do you mean that if it's true that the sky is blue then it's blue whatever else is? what's the phrase for that?

    – another_name
    9 hours ago











  • @another_name I put the unicorns in there to give an example of an argument that might be considered nonsense. It could be any statement.

    – Frank Hubeny
    9 hours ago











  • sure, i'm not complaining about the 'unicorn'. i'm just still not sure what it means to say that B > [A > B]. does it have a name?

    – another_name
    9 hours ago












  • @another_name It could be thought of as reiteration. One could remove the A entirely and write B > B. It could be viewed as a circular argument or begging the question. This would be fallacious unless all sides in an argument accepted B.

    – Frank Hubeny
    7 hours ago



















4














The topic you want to research is 'paradoxes of material implication.' That is, you are right to think there's something odd about this formula. But it does not mean that every fact deductively follows from every other. Consider your target string:




B-->(A-->B)




Now note that the convention is to define the connectives such that the conditional 'P-->Q' is logically equivalent to '~PvQ'. Hence, replacing each instance of the conditional with the disjunctive equivalent, your target string is logically equivalent to:




~B v (~AvB).




i.e., Either not-B, or else not-A or B. On this formulation, it is clear that this formula does not mean what you suggest it means. Furthermore, this disjunction is a trivial logical truth . There are two ways the world could be with respect to B: either B is true or it is not true. If B is true, this formula is satisfied by the second disjunct. If B is not true, this formula is satisfied by the first disjunct. Hence, it is logically true.



This formula is one among many reasons why philosophers have thought that the material conditional of first-order logic does not capture any interesting epistemically-rich notion of implication. When formulated as a conditional, this leads to valid implications that are counterintuitive, pragmatically odd, or prehaps even false. That is, we usually expect implications to carry some kind of epistemic or pragmatic force, to deliver some information beyond the initial terms. So carrying out this kind of inference, though valid, holds no water in ordinary discourse. Depending on your views about the relationship between semantics and pragmatics, then, there is actually an argument to be made that such "implications" are a kind of nonsense.



Monotonicity is another feature of logic that you may want to look into. Entailment in first-order logic is a monotonic operation, which in this setting means, intuitively, that once something B is proven, the addition of new assumptions will never make B false. One way to think about your formula is through the monotonicity of entailment. Suppose we have some string of background assumptions T, such that T entails B:




T |- B




The monotonicity of entailment ensures that the addition of new assumptions will never prohibit concluding B. Hence, the extension of T by arbitrary new assumption A will not alter the result:




T, A |- B




Many systems are intuitively non-monotonic, such as when we reason in a default way or when we assume that implications should have some relevance to one another. If you're in the United States, then you do default reasoning whenever you pay your taxes. For example, by default you owe the standard amount of taxes for your income level. But the addition of other assumptions --- such as itemized deductions, being the owner of a business, the head of a household, of retirement age, capital gains, student loan interest payments, etc. --- can make those default truths false. We also usually demand that the terms of an inference be relevant to one another. This is, at least arguably, one point of departure between traditional Aristotelian logic and modern mathematical logic. Aristotle's logic exhibits features of a relevance logic, i.e., the assumption that the premises in an argument must have some relevance to one another by virtue of which the terms of the premises can be combined or separated. This is one of the many ways in whcih Aristotle's understanding of syllogisms and deduction differs from our modern notions of validity. How to pin down 'relevance' is an entirely other can of worms. In a Wittgensteinian vein, we might also complain that someone who asserted a conditional like yours in conversation would be failing to make a move in the language game.



First-order logic validates all kinds of at best pragmatically useless and---again, depending on your background philosophical views---perhaps false entailments.



Consider:
Priest, Graham (2001). Introduction to Non-Classical Logic. Cambridge University Press, pp.12-13 (on material implication)



https://projecteuclid.org/download/pdf_1/euclid.ndjfl/1093883397



https://en.wikipedia.org/wiki/Monotonicity_of_entailment



https://plato.stanford.edu/entries/logic-relevance/



https://plato.stanford.edu/entries/logic-nonmonotonic/






share|improve this answer
































    2














    I take a statement to be ridiculous iff it contains or implies a contradiction. That being said, B --> (A-->B) is not ridiculous, but it is a tautology, which makes it vacuous.



    I feel the previous answers over complicated things, but here are two short arguments showing that it is a tautology.



    1. B --> (A --> B) Assumption 
    2. B --> (~A v B) 1, Implication
    3. ~B v (~A v B) 2, Implication
    4. ~B v (B v ~A) 3, Commutation
    5. (~B v B) v ~A 4, Association


    You get the same solution using the rule of Exportation



    1. B --> (A --> B) Assumption
    2. (B ^ A) --> B 1, Exportation
    3. ~(B ^ A) v B 2, Implication
    4. (~B v ~A) v B 3, De Morgan
    5. (~A v ~B) v B 4, Commutation
    6. ~A v (~B v B) 5, Association
    7. (~B v B) v ~A 6, Commutation


    ~B v B is logically true. It is trivial to show that (~B v ~B) v ~A is also logically true.



    (@Frank Hubeny Thanks for keeping me honest and reminding me to include the Commutation steps in both proofs).






    share|improve this answer










    New contributor



    Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.



















    • There is also a commutative operation after line 3 in the first proof:. (~A v B) rewritten as (B v ~A). In the second, shouldn't the second line be ~B v (A --> B)? Regardless, this is another way to show the result. +1

      – Frank Hubeny
      3 hours ago












    • You are correct, I did skip the commutation move in the first proof. I'll fix that. In the second proof at line 2 I used a rule called Exportation which says P--> (Q-->R) <---> (P ^ Q)-->R. Thanks for the vote!

      – Rob
      3 hours ago



















    1














    It is a tautology.



    •If B is true, then A->B is true. Then B->(A->B) is B->T. B->T is T.



    •If B is false, then B->(A->B) is F->(A->B). F-> Something is always True. That's why B->(A->B) is true.






    share|improve this answer








    New contributor



    Asım Kaya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.



















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      4 Answers
      4






      active

      oldest

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      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      When proving a conditional one assumes the antecedent, B. The goal is not to derive this, but from this assumption to derive the consequent which happens to be A > B.



      But A > B is another conditional. Since it is a conditional one derives that in the same way. First assume the antecedent, A. Can one derive B? Yes, one can, because in this derivation we already assumed B in the first line. We can "reiterate" it in the next line as B.



      Since from A one can derive B in a subproof, one can introduce a conditional (→I), A > B. This is the same as saying: if I have A, I can derive B. It is a shorthand for that subproof.



      But if one has A > B one can introduce the conditional again and write B > [A > B].



      Here's the result using the proof checker linked to below:



      enter image description here



      What this shows is that the argument is valid. The argument might be nonsense depending on what A and B are. Suppose A is Unicorns are white and B is The sky is blue. What this says is If the sky is blue then if unicorns are white then the sky is blue. Suppose the sky is blue. It doesn't matter whether unicorns are white or not. The sky is still blue.



      For the reiteration rule (R) see section 16.1 of the forallx text below.




      Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



      P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/






      share|improve this answer























      • thanks, i think i followed the answer, though it's not really in terms i feel comfortable with. you did lose me a bit when talking about unicorns. do you mean that if it's true that the sky is blue then it's blue whatever else is? what's the phrase for that?

        – another_name
        9 hours ago











      • @another_name I put the unicorns in there to give an example of an argument that might be considered nonsense. It could be any statement.

        – Frank Hubeny
        9 hours ago











      • sure, i'm not complaining about the 'unicorn'. i'm just still not sure what it means to say that B > [A > B]. does it have a name?

        – another_name
        9 hours ago












      • @another_name It could be thought of as reiteration. One could remove the A entirely and write B > B. It could be viewed as a circular argument or begging the question. This would be fallacious unless all sides in an argument accepted B.

        – Frank Hubeny
        7 hours ago
















      4














      When proving a conditional one assumes the antecedent, B. The goal is not to derive this, but from this assumption to derive the consequent which happens to be A > B.



      But A > B is another conditional. Since it is a conditional one derives that in the same way. First assume the antecedent, A. Can one derive B? Yes, one can, because in this derivation we already assumed B in the first line. We can "reiterate" it in the next line as B.



      Since from A one can derive B in a subproof, one can introduce a conditional (→I), A > B. This is the same as saying: if I have A, I can derive B. It is a shorthand for that subproof.



      But if one has A > B one can introduce the conditional again and write B > [A > B].



      Here's the result using the proof checker linked to below:



      enter image description here



      What this shows is that the argument is valid. The argument might be nonsense depending on what A and B are. Suppose A is Unicorns are white and B is The sky is blue. What this says is If the sky is blue then if unicorns are white then the sky is blue. Suppose the sky is blue. It doesn't matter whether unicorns are white or not. The sky is still blue.



      For the reiteration rule (R) see section 16.1 of the forallx text below.




      Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



      P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/






      share|improve this answer























      • thanks, i think i followed the answer, though it's not really in terms i feel comfortable with. you did lose me a bit when talking about unicorns. do you mean that if it's true that the sky is blue then it's blue whatever else is? what's the phrase for that?

        – another_name
        9 hours ago











      • @another_name I put the unicorns in there to give an example of an argument that might be considered nonsense. It could be any statement.

        – Frank Hubeny
        9 hours ago











      • sure, i'm not complaining about the 'unicorn'. i'm just still not sure what it means to say that B > [A > B]. does it have a name?

        – another_name
        9 hours ago












      • @another_name It could be thought of as reiteration. One could remove the A entirely and write B > B. It could be viewed as a circular argument or begging the question. This would be fallacious unless all sides in an argument accepted B.

        – Frank Hubeny
        7 hours ago














      4












      4








      4







      When proving a conditional one assumes the antecedent, B. The goal is not to derive this, but from this assumption to derive the consequent which happens to be A > B.



      But A > B is another conditional. Since it is a conditional one derives that in the same way. First assume the antecedent, A. Can one derive B? Yes, one can, because in this derivation we already assumed B in the first line. We can "reiterate" it in the next line as B.



      Since from A one can derive B in a subproof, one can introduce a conditional (→I), A > B. This is the same as saying: if I have A, I can derive B. It is a shorthand for that subproof.



      But if one has A > B one can introduce the conditional again and write B > [A > B].



      Here's the result using the proof checker linked to below:



      enter image description here



      What this shows is that the argument is valid. The argument might be nonsense depending on what A and B are. Suppose A is Unicorns are white and B is The sky is blue. What this says is If the sky is blue then if unicorns are white then the sky is blue. Suppose the sky is blue. It doesn't matter whether unicorns are white or not. The sky is still blue.



      For the reiteration rule (R) see section 16.1 of the forallx text below.




      Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



      P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/






      share|improve this answer













      When proving a conditional one assumes the antecedent, B. The goal is not to derive this, but from this assumption to derive the consequent which happens to be A > B.



      But A > B is another conditional. Since it is a conditional one derives that in the same way. First assume the antecedent, A. Can one derive B? Yes, one can, because in this derivation we already assumed B in the first line. We can "reiterate" it in the next line as B.



      Since from A one can derive B in a subproof, one can introduce a conditional (→I), A > B. This is the same as saying: if I have A, I can derive B. It is a shorthand for that subproof.



      But if one has A > B one can introduce the conditional again and write B > [A > B].



      Here's the result using the proof checker linked to below:



      enter image description here



      What this shows is that the argument is valid. The argument might be nonsense depending on what A and B are. Suppose A is Unicorns are white and B is The sky is blue. What this says is If the sky is blue then if unicorns are white then the sky is blue. Suppose the sky is blue. It doesn't matter whether unicorns are white or not. The sky is still blue.



      For the reiteration rule (R) see section 16.1 of the forallx text below.




      Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



      P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 9 hours ago









      Frank HubenyFrank Hubeny

      11.1k51559




      11.1k51559












      • thanks, i think i followed the answer, though it's not really in terms i feel comfortable with. you did lose me a bit when talking about unicorns. do you mean that if it's true that the sky is blue then it's blue whatever else is? what's the phrase for that?

        – another_name
        9 hours ago











      • @another_name I put the unicorns in there to give an example of an argument that might be considered nonsense. It could be any statement.

        – Frank Hubeny
        9 hours ago











      • sure, i'm not complaining about the 'unicorn'. i'm just still not sure what it means to say that B > [A > B]. does it have a name?

        – another_name
        9 hours ago












      • @another_name It could be thought of as reiteration. One could remove the A entirely and write B > B. It could be viewed as a circular argument or begging the question. This would be fallacious unless all sides in an argument accepted B.

        – Frank Hubeny
        7 hours ago


















      • thanks, i think i followed the answer, though it's not really in terms i feel comfortable with. you did lose me a bit when talking about unicorns. do you mean that if it's true that the sky is blue then it's blue whatever else is? what's the phrase for that?

        – another_name
        9 hours ago











      • @another_name I put the unicorns in there to give an example of an argument that might be considered nonsense. It could be any statement.

        – Frank Hubeny
        9 hours ago











      • sure, i'm not complaining about the 'unicorn'. i'm just still not sure what it means to say that B > [A > B]. does it have a name?

        – another_name
        9 hours ago












      • @another_name It could be thought of as reiteration. One could remove the A entirely and write B > B. It could be viewed as a circular argument or begging the question. This would be fallacious unless all sides in an argument accepted B.

        – Frank Hubeny
        7 hours ago

















      thanks, i think i followed the answer, though it's not really in terms i feel comfortable with. you did lose me a bit when talking about unicorns. do you mean that if it's true that the sky is blue then it's blue whatever else is? what's the phrase for that?

      – another_name
      9 hours ago





      thanks, i think i followed the answer, though it's not really in terms i feel comfortable with. you did lose me a bit when talking about unicorns. do you mean that if it's true that the sky is blue then it's blue whatever else is? what's the phrase for that?

      – another_name
      9 hours ago













      @another_name I put the unicorns in there to give an example of an argument that might be considered nonsense. It could be any statement.

      – Frank Hubeny
      9 hours ago





      @another_name I put the unicorns in there to give an example of an argument that might be considered nonsense. It could be any statement.

      – Frank Hubeny
      9 hours ago













      sure, i'm not complaining about the 'unicorn'. i'm just still not sure what it means to say that B > [A > B]. does it have a name?

      – another_name
      9 hours ago






      sure, i'm not complaining about the 'unicorn'. i'm just still not sure what it means to say that B > [A > B]. does it have a name?

      – another_name
      9 hours ago














      @another_name It could be thought of as reiteration. One could remove the A entirely and write B > B. It could be viewed as a circular argument or begging the question. This would be fallacious unless all sides in an argument accepted B.

      – Frank Hubeny
      7 hours ago






      @another_name It could be thought of as reiteration. One could remove the A entirely and write B > B. It could be viewed as a circular argument or begging the question. This would be fallacious unless all sides in an argument accepted B.

      – Frank Hubeny
      7 hours ago












      4














      The topic you want to research is 'paradoxes of material implication.' That is, you are right to think there's something odd about this formula. But it does not mean that every fact deductively follows from every other. Consider your target string:




      B-->(A-->B)




      Now note that the convention is to define the connectives such that the conditional 'P-->Q' is logically equivalent to '~PvQ'. Hence, replacing each instance of the conditional with the disjunctive equivalent, your target string is logically equivalent to:




      ~B v (~AvB).




      i.e., Either not-B, or else not-A or B. On this formulation, it is clear that this formula does not mean what you suggest it means. Furthermore, this disjunction is a trivial logical truth . There are two ways the world could be with respect to B: either B is true or it is not true. If B is true, this formula is satisfied by the second disjunct. If B is not true, this formula is satisfied by the first disjunct. Hence, it is logically true.



      This formula is one among many reasons why philosophers have thought that the material conditional of first-order logic does not capture any interesting epistemically-rich notion of implication. When formulated as a conditional, this leads to valid implications that are counterintuitive, pragmatically odd, or prehaps even false. That is, we usually expect implications to carry some kind of epistemic or pragmatic force, to deliver some information beyond the initial terms. So carrying out this kind of inference, though valid, holds no water in ordinary discourse. Depending on your views about the relationship between semantics and pragmatics, then, there is actually an argument to be made that such "implications" are a kind of nonsense.



      Monotonicity is another feature of logic that you may want to look into. Entailment in first-order logic is a monotonic operation, which in this setting means, intuitively, that once something B is proven, the addition of new assumptions will never make B false. One way to think about your formula is through the monotonicity of entailment. Suppose we have some string of background assumptions T, such that T entails B:




      T |- B




      The monotonicity of entailment ensures that the addition of new assumptions will never prohibit concluding B. Hence, the extension of T by arbitrary new assumption A will not alter the result:




      T, A |- B




      Many systems are intuitively non-monotonic, such as when we reason in a default way or when we assume that implications should have some relevance to one another. If you're in the United States, then you do default reasoning whenever you pay your taxes. For example, by default you owe the standard amount of taxes for your income level. But the addition of other assumptions --- such as itemized deductions, being the owner of a business, the head of a household, of retirement age, capital gains, student loan interest payments, etc. --- can make those default truths false. We also usually demand that the terms of an inference be relevant to one another. This is, at least arguably, one point of departure between traditional Aristotelian logic and modern mathematical logic. Aristotle's logic exhibits features of a relevance logic, i.e., the assumption that the premises in an argument must have some relevance to one another by virtue of which the terms of the premises can be combined or separated. This is one of the many ways in whcih Aristotle's understanding of syllogisms and deduction differs from our modern notions of validity. How to pin down 'relevance' is an entirely other can of worms. In a Wittgensteinian vein, we might also complain that someone who asserted a conditional like yours in conversation would be failing to make a move in the language game.



      First-order logic validates all kinds of at best pragmatically useless and---again, depending on your background philosophical views---perhaps false entailments.



      Consider:
      Priest, Graham (2001). Introduction to Non-Classical Logic. Cambridge University Press, pp.12-13 (on material implication)



      https://projecteuclid.org/download/pdf_1/euclid.ndjfl/1093883397



      https://en.wikipedia.org/wiki/Monotonicity_of_entailment



      https://plato.stanford.edu/entries/logic-relevance/



      https://plato.stanford.edu/entries/logic-nonmonotonic/






      share|improve this answer





























        4














        The topic you want to research is 'paradoxes of material implication.' That is, you are right to think there's something odd about this formula. But it does not mean that every fact deductively follows from every other. Consider your target string:




        B-->(A-->B)




        Now note that the convention is to define the connectives such that the conditional 'P-->Q' is logically equivalent to '~PvQ'. Hence, replacing each instance of the conditional with the disjunctive equivalent, your target string is logically equivalent to:




        ~B v (~AvB).




        i.e., Either not-B, or else not-A or B. On this formulation, it is clear that this formula does not mean what you suggest it means. Furthermore, this disjunction is a trivial logical truth . There are two ways the world could be with respect to B: either B is true or it is not true. If B is true, this formula is satisfied by the second disjunct. If B is not true, this formula is satisfied by the first disjunct. Hence, it is logically true.



        This formula is one among many reasons why philosophers have thought that the material conditional of first-order logic does not capture any interesting epistemically-rich notion of implication. When formulated as a conditional, this leads to valid implications that are counterintuitive, pragmatically odd, or prehaps even false. That is, we usually expect implications to carry some kind of epistemic or pragmatic force, to deliver some information beyond the initial terms. So carrying out this kind of inference, though valid, holds no water in ordinary discourse. Depending on your views about the relationship between semantics and pragmatics, then, there is actually an argument to be made that such "implications" are a kind of nonsense.



        Monotonicity is another feature of logic that you may want to look into. Entailment in first-order logic is a monotonic operation, which in this setting means, intuitively, that once something B is proven, the addition of new assumptions will never make B false. One way to think about your formula is through the monotonicity of entailment. Suppose we have some string of background assumptions T, such that T entails B:




        T |- B




        The monotonicity of entailment ensures that the addition of new assumptions will never prohibit concluding B. Hence, the extension of T by arbitrary new assumption A will not alter the result:




        T, A |- B




        Many systems are intuitively non-monotonic, such as when we reason in a default way or when we assume that implications should have some relevance to one another. If you're in the United States, then you do default reasoning whenever you pay your taxes. For example, by default you owe the standard amount of taxes for your income level. But the addition of other assumptions --- such as itemized deductions, being the owner of a business, the head of a household, of retirement age, capital gains, student loan interest payments, etc. --- can make those default truths false. We also usually demand that the terms of an inference be relevant to one another. This is, at least arguably, one point of departure between traditional Aristotelian logic and modern mathematical logic. Aristotle's logic exhibits features of a relevance logic, i.e., the assumption that the premises in an argument must have some relevance to one another by virtue of which the terms of the premises can be combined or separated. This is one of the many ways in whcih Aristotle's understanding of syllogisms and deduction differs from our modern notions of validity. How to pin down 'relevance' is an entirely other can of worms. In a Wittgensteinian vein, we might also complain that someone who asserted a conditional like yours in conversation would be failing to make a move in the language game.



        First-order logic validates all kinds of at best pragmatically useless and---again, depending on your background philosophical views---perhaps false entailments.



        Consider:
        Priest, Graham (2001). Introduction to Non-Classical Logic. Cambridge University Press, pp.12-13 (on material implication)



        https://projecteuclid.org/download/pdf_1/euclid.ndjfl/1093883397



        https://en.wikipedia.org/wiki/Monotonicity_of_entailment



        https://plato.stanford.edu/entries/logic-relevance/



        https://plato.stanford.edu/entries/logic-nonmonotonic/






        share|improve this answer



























          4












          4








          4







          The topic you want to research is 'paradoxes of material implication.' That is, you are right to think there's something odd about this formula. But it does not mean that every fact deductively follows from every other. Consider your target string:




          B-->(A-->B)




          Now note that the convention is to define the connectives such that the conditional 'P-->Q' is logically equivalent to '~PvQ'. Hence, replacing each instance of the conditional with the disjunctive equivalent, your target string is logically equivalent to:




          ~B v (~AvB).




          i.e., Either not-B, or else not-A or B. On this formulation, it is clear that this formula does not mean what you suggest it means. Furthermore, this disjunction is a trivial logical truth . There are two ways the world could be with respect to B: either B is true or it is not true. If B is true, this formula is satisfied by the second disjunct. If B is not true, this formula is satisfied by the first disjunct. Hence, it is logically true.



          This formula is one among many reasons why philosophers have thought that the material conditional of first-order logic does not capture any interesting epistemically-rich notion of implication. When formulated as a conditional, this leads to valid implications that are counterintuitive, pragmatically odd, or prehaps even false. That is, we usually expect implications to carry some kind of epistemic or pragmatic force, to deliver some information beyond the initial terms. So carrying out this kind of inference, though valid, holds no water in ordinary discourse. Depending on your views about the relationship between semantics and pragmatics, then, there is actually an argument to be made that such "implications" are a kind of nonsense.



          Monotonicity is another feature of logic that you may want to look into. Entailment in first-order logic is a monotonic operation, which in this setting means, intuitively, that once something B is proven, the addition of new assumptions will never make B false. One way to think about your formula is through the monotonicity of entailment. Suppose we have some string of background assumptions T, such that T entails B:




          T |- B




          The monotonicity of entailment ensures that the addition of new assumptions will never prohibit concluding B. Hence, the extension of T by arbitrary new assumption A will not alter the result:




          T, A |- B




          Many systems are intuitively non-monotonic, such as when we reason in a default way or when we assume that implications should have some relevance to one another. If you're in the United States, then you do default reasoning whenever you pay your taxes. For example, by default you owe the standard amount of taxes for your income level. But the addition of other assumptions --- such as itemized deductions, being the owner of a business, the head of a household, of retirement age, capital gains, student loan interest payments, etc. --- can make those default truths false. We also usually demand that the terms of an inference be relevant to one another. This is, at least arguably, one point of departure between traditional Aristotelian logic and modern mathematical logic. Aristotle's logic exhibits features of a relevance logic, i.e., the assumption that the premises in an argument must have some relevance to one another by virtue of which the terms of the premises can be combined or separated. This is one of the many ways in whcih Aristotle's understanding of syllogisms and deduction differs from our modern notions of validity. How to pin down 'relevance' is an entirely other can of worms. In a Wittgensteinian vein, we might also complain that someone who asserted a conditional like yours in conversation would be failing to make a move in the language game.



          First-order logic validates all kinds of at best pragmatically useless and---again, depending on your background philosophical views---perhaps false entailments.



          Consider:
          Priest, Graham (2001). Introduction to Non-Classical Logic. Cambridge University Press, pp.12-13 (on material implication)



          https://projecteuclid.org/download/pdf_1/euclid.ndjfl/1093883397



          https://en.wikipedia.org/wiki/Monotonicity_of_entailment



          https://plato.stanford.edu/entries/logic-relevance/



          https://plato.stanford.edu/entries/logic-nonmonotonic/






          share|improve this answer















          The topic you want to research is 'paradoxes of material implication.' That is, you are right to think there's something odd about this formula. But it does not mean that every fact deductively follows from every other. Consider your target string:




          B-->(A-->B)




          Now note that the convention is to define the connectives such that the conditional 'P-->Q' is logically equivalent to '~PvQ'. Hence, replacing each instance of the conditional with the disjunctive equivalent, your target string is logically equivalent to:




          ~B v (~AvB).




          i.e., Either not-B, or else not-A or B. On this formulation, it is clear that this formula does not mean what you suggest it means. Furthermore, this disjunction is a trivial logical truth . There are two ways the world could be with respect to B: either B is true or it is not true. If B is true, this formula is satisfied by the second disjunct. If B is not true, this formula is satisfied by the first disjunct. Hence, it is logically true.



          This formula is one among many reasons why philosophers have thought that the material conditional of first-order logic does not capture any interesting epistemically-rich notion of implication. When formulated as a conditional, this leads to valid implications that are counterintuitive, pragmatically odd, or prehaps even false. That is, we usually expect implications to carry some kind of epistemic or pragmatic force, to deliver some information beyond the initial terms. So carrying out this kind of inference, though valid, holds no water in ordinary discourse. Depending on your views about the relationship between semantics and pragmatics, then, there is actually an argument to be made that such "implications" are a kind of nonsense.



          Monotonicity is another feature of logic that you may want to look into. Entailment in first-order logic is a monotonic operation, which in this setting means, intuitively, that once something B is proven, the addition of new assumptions will never make B false. One way to think about your formula is through the monotonicity of entailment. Suppose we have some string of background assumptions T, such that T entails B:




          T |- B




          The monotonicity of entailment ensures that the addition of new assumptions will never prohibit concluding B. Hence, the extension of T by arbitrary new assumption A will not alter the result:




          T, A |- B




          Many systems are intuitively non-monotonic, such as when we reason in a default way or when we assume that implications should have some relevance to one another. If you're in the United States, then you do default reasoning whenever you pay your taxes. For example, by default you owe the standard amount of taxes for your income level. But the addition of other assumptions --- such as itemized deductions, being the owner of a business, the head of a household, of retirement age, capital gains, student loan interest payments, etc. --- can make those default truths false. We also usually demand that the terms of an inference be relevant to one another. This is, at least arguably, one point of departure between traditional Aristotelian logic and modern mathematical logic. Aristotle's logic exhibits features of a relevance logic, i.e., the assumption that the premises in an argument must have some relevance to one another by virtue of which the terms of the premises can be combined or separated. This is one of the many ways in whcih Aristotle's understanding of syllogisms and deduction differs from our modern notions of validity. How to pin down 'relevance' is an entirely other can of worms. In a Wittgensteinian vein, we might also complain that someone who asserted a conditional like yours in conversation would be failing to make a move in the language game.



          First-order logic validates all kinds of at best pragmatically useless and---again, depending on your background philosophical views---perhaps false entailments.



          Consider:
          Priest, Graham (2001). Introduction to Non-Classical Logic. Cambridge University Press, pp.12-13 (on material implication)



          https://projecteuclid.org/download/pdf_1/euclid.ndjfl/1093883397



          https://en.wikipedia.org/wiki/Monotonicity_of_entailment



          https://plato.stanford.edu/entries/logic-relevance/



          https://plato.stanford.edu/entries/logic-nonmonotonic/







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 6 hours ago

























          answered 7 hours ago









          transitionsynthesistransitionsynthesis

          94867




          94867





















              2














              I take a statement to be ridiculous iff it contains or implies a contradiction. That being said, B --> (A-->B) is not ridiculous, but it is a tautology, which makes it vacuous.



              I feel the previous answers over complicated things, but here are two short arguments showing that it is a tautology.



              1. B --> (A --> B) Assumption 
              2. B --> (~A v B) 1, Implication
              3. ~B v (~A v B) 2, Implication
              4. ~B v (B v ~A) 3, Commutation
              5. (~B v B) v ~A 4, Association


              You get the same solution using the rule of Exportation



              1. B --> (A --> B) Assumption
              2. (B ^ A) --> B 1, Exportation
              3. ~(B ^ A) v B 2, Implication
              4. (~B v ~A) v B 3, De Morgan
              5. (~A v ~B) v B 4, Commutation
              6. ~A v (~B v B) 5, Association
              7. (~B v B) v ~A 6, Commutation


              ~B v B is logically true. It is trivial to show that (~B v ~B) v ~A is also logically true.



              (@Frank Hubeny Thanks for keeping me honest and reminding me to include the Commutation steps in both proofs).






              share|improve this answer










              New contributor



              Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.



















              • There is also a commutative operation after line 3 in the first proof:. (~A v B) rewritten as (B v ~A). In the second, shouldn't the second line be ~B v (A --> B)? Regardless, this is another way to show the result. +1

                – Frank Hubeny
                3 hours ago












              • You are correct, I did skip the commutation move in the first proof. I'll fix that. In the second proof at line 2 I used a rule called Exportation which says P--> (Q-->R) <---> (P ^ Q)-->R. Thanks for the vote!

                – Rob
                3 hours ago
















              2














              I take a statement to be ridiculous iff it contains or implies a contradiction. That being said, B --> (A-->B) is not ridiculous, but it is a tautology, which makes it vacuous.



              I feel the previous answers over complicated things, but here are two short arguments showing that it is a tautology.



              1. B --> (A --> B) Assumption 
              2. B --> (~A v B) 1, Implication
              3. ~B v (~A v B) 2, Implication
              4. ~B v (B v ~A) 3, Commutation
              5. (~B v B) v ~A 4, Association


              You get the same solution using the rule of Exportation



              1. B --> (A --> B) Assumption
              2. (B ^ A) --> B 1, Exportation
              3. ~(B ^ A) v B 2, Implication
              4. (~B v ~A) v B 3, De Morgan
              5. (~A v ~B) v B 4, Commutation
              6. ~A v (~B v B) 5, Association
              7. (~B v B) v ~A 6, Commutation


              ~B v B is logically true. It is trivial to show that (~B v ~B) v ~A is also logically true.



              (@Frank Hubeny Thanks for keeping me honest and reminding me to include the Commutation steps in both proofs).






              share|improve this answer










              New contributor



              Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.



















              • There is also a commutative operation after line 3 in the first proof:. (~A v B) rewritten as (B v ~A). In the second, shouldn't the second line be ~B v (A --> B)? Regardless, this is another way to show the result. +1

                – Frank Hubeny
                3 hours ago












              • You are correct, I did skip the commutation move in the first proof. I'll fix that. In the second proof at line 2 I used a rule called Exportation which says P--> (Q-->R) <---> (P ^ Q)-->R. Thanks for the vote!

                – Rob
                3 hours ago














              2












              2








              2







              I take a statement to be ridiculous iff it contains or implies a contradiction. That being said, B --> (A-->B) is not ridiculous, but it is a tautology, which makes it vacuous.



              I feel the previous answers over complicated things, but here are two short arguments showing that it is a tautology.



              1. B --> (A --> B) Assumption 
              2. B --> (~A v B) 1, Implication
              3. ~B v (~A v B) 2, Implication
              4. ~B v (B v ~A) 3, Commutation
              5. (~B v B) v ~A 4, Association


              You get the same solution using the rule of Exportation



              1. B --> (A --> B) Assumption
              2. (B ^ A) --> B 1, Exportation
              3. ~(B ^ A) v B 2, Implication
              4. (~B v ~A) v B 3, De Morgan
              5. (~A v ~B) v B 4, Commutation
              6. ~A v (~B v B) 5, Association
              7. (~B v B) v ~A 6, Commutation


              ~B v B is logically true. It is trivial to show that (~B v ~B) v ~A is also logically true.



              (@Frank Hubeny Thanks for keeping me honest and reminding me to include the Commutation steps in both proofs).






              share|improve this answer










              New contributor



              Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.









              I take a statement to be ridiculous iff it contains or implies a contradiction. That being said, B --> (A-->B) is not ridiculous, but it is a tautology, which makes it vacuous.



              I feel the previous answers over complicated things, but here are two short arguments showing that it is a tautology.



              1. B --> (A --> B) Assumption 
              2. B --> (~A v B) 1, Implication
              3. ~B v (~A v B) 2, Implication
              4. ~B v (B v ~A) 3, Commutation
              5. (~B v B) v ~A 4, Association


              You get the same solution using the rule of Exportation



              1. B --> (A --> B) Assumption
              2. (B ^ A) --> B 1, Exportation
              3. ~(B ^ A) v B 2, Implication
              4. (~B v ~A) v B 3, De Morgan
              5. (~A v ~B) v B 4, Commutation
              6. ~A v (~B v B) 5, Association
              7. (~B v B) v ~A 6, Commutation


              ~B v B is logically true. It is trivial to show that (~B v ~B) v ~A is also logically true.



              (@Frank Hubeny Thanks for keeping me honest and reminding me to include the Commutation steps in both proofs).







              share|improve this answer










              New contributor



              Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.








              share|improve this answer



              share|improve this answer








              edited 3 hours ago





















              New contributor



              Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.








              answered 3 hours ago









              RobRob

              212




              212




              New contributor



              Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.




              New contributor




              Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.














              • There is also a commutative operation after line 3 in the first proof:. (~A v B) rewritten as (B v ~A). In the second, shouldn't the second line be ~B v (A --> B)? Regardless, this is another way to show the result. +1

                – Frank Hubeny
                3 hours ago












              • You are correct, I did skip the commutation move in the first proof. I'll fix that. In the second proof at line 2 I used a rule called Exportation which says P--> (Q-->R) <---> (P ^ Q)-->R. Thanks for the vote!

                – Rob
                3 hours ago


















              • There is also a commutative operation after line 3 in the first proof:. (~A v B) rewritten as (B v ~A). In the second, shouldn't the second line be ~B v (A --> B)? Regardless, this is another way to show the result. +1

                – Frank Hubeny
                3 hours ago












              • You are correct, I did skip the commutation move in the first proof. I'll fix that. In the second proof at line 2 I used a rule called Exportation which says P--> (Q-->R) <---> (P ^ Q)-->R. Thanks for the vote!

                – Rob
                3 hours ago

















              There is also a commutative operation after line 3 in the first proof:. (~A v B) rewritten as (B v ~A). In the second, shouldn't the second line be ~B v (A --> B)? Regardless, this is another way to show the result. +1

              – Frank Hubeny
              3 hours ago






              There is also a commutative operation after line 3 in the first proof:. (~A v B) rewritten as (B v ~A). In the second, shouldn't the second line be ~B v (A --> B)? Regardless, this is another way to show the result. +1

              – Frank Hubeny
              3 hours ago














              You are correct, I did skip the commutation move in the first proof. I'll fix that. In the second proof at line 2 I used a rule called Exportation which says P--> (Q-->R) <---> (P ^ Q)-->R. Thanks for the vote!

              – Rob
              3 hours ago






              You are correct, I did skip the commutation move in the first proof. I'll fix that. In the second proof at line 2 I used a rule called Exportation which says P--> (Q-->R) <---> (P ^ Q)-->R. Thanks for the vote!

              – Rob
              3 hours ago












              1














              It is a tautology.



              •If B is true, then A->B is true. Then B->(A->B) is B->T. B->T is T.



              •If B is false, then B->(A->B) is F->(A->B). F-> Something is always True. That's why B->(A->B) is true.






              share|improve this answer








              New contributor



              Asım Kaya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.























                1














                It is a tautology.



                •If B is true, then A->B is true. Then B->(A->B) is B->T. B->T is T.



                •If B is false, then B->(A->B) is F->(A->B). F-> Something is always True. That's why B->(A->B) is true.






                share|improve this answer








                New contributor



                Asım Kaya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





















                  1












                  1








                  1







                  It is a tautology.



                  •If B is true, then A->B is true. Then B->(A->B) is B->T. B->T is T.



                  •If B is false, then B->(A->B) is F->(A->B). F-> Something is always True. That's why B->(A->B) is true.






                  share|improve this answer








                  New contributor



                  Asım Kaya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  It is a tautology.



                  •If B is true, then A->B is true. Then B->(A->B) is B->T. B->T is T.



                  •If B is false, then B->(A->B) is F->(A->B). F-> Something is always True. That's why B->(A->B) is true.







                  share|improve this answer








                  New contributor



                  Asım Kaya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.








                  share|improve this answer



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