Etydhv

How wide is a neg symbol, how to get the width for alignment?

Can an isometry leave entropy invariant?

Prove that the limit exists or does not exist

How to model the curly cable part of the phone

How long would it take for people to notice a mass disappearance?

What to use instead of cling film to wrap pastry

If your medical expenses exceed your income does the IRS pay you?

What property of a BJT transistor makes it an amplifier?

Why is the relative clause in the following sentence not directly after the noun and why is the verb not in the end of the sentence?

Independent, post-Brexit Scotland - would there be a hard border with England?

What is the most remote airport from the center of the city it supposedly serves?

String won't reverse using reverse_copy

I need a disease

Can a nothic's Weird Insight action discover secrets about a player character that the character doesn't know about themselves?

How do LIGO and VIRGO know that a gravitational wave has its origin in a neutron star or a black hole?

Multi-channel audio upsampling interpolation

Why wasn't the Night King naked in S08E03?

Can you complete the sequence?

Why do money exchangers give different rates to different bills?

What are the differences between credential stuffing and password spraying?

I drew a randomly colored grid of points with tikz, how do I force it to remember the first grid from then on?

What happens if you dump antimatter into a black hole?

Can hackers enable the camera after the user disabled it?

Where can I go to avoid planes overhead?

Fitch Proof Question

Conditional disjunction equivalence proof using FItchFitch Proof - Logic LPL 13.11Fitch Proof - LPL Exercise 8.17fitch proof chapter 13 exercise 13.49Use the Fitch system to prove the tautology (p ∨ ¬p)LPL ( language proof and logic) - FITCH - 14.12Seminal, accessible work on defeasible reasoningIs it possible to define argument validity as a formula?Fitch Biconditional Proof Help?Fitch Proof by Contradiction help

1

I'm having trouble with a proof and I'm not sure if it's valid or not. If it appears to be invalid, we are supposed to assign names to the letters in the proof and check it in a World, but when I do it still checks out.

When I try to prove it in Fitch, I get all the way to the bottom, with (Q^R) as the last step of the subproof and I'm trying to apply -->Intro to join
(NOT)P --> (Q ^ R) but it's not checking out because it says that it needs a support step to be cited.

P --> Q Premise
R ^ S Premise
------------------
(NOT)P --> (Q ^ R) Goal

logic proof language fitch

share|improve this question

edited 1 hour ago

Frank Hubeny

11k51559

asked 2 hours ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Nope, that is not valid. Are you missing a 'not' before that first P?
  
  – Graham Kemp
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is it not valid because there's no (NOT)P in the premises? Since it's not known? No, it's not missing. The first premise is P --> Q.
  
  – fitch-is-killing-me
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Then you cannot conclude ¬P → (Q ˄ R) from those premises, although you may conclude P → (Q ˄ R)
  
  – Graham Kemp
  2 hours ago
  

  
  
  
  

add a comment | 

1

I'm having trouble with a proof and I'm not sure if it's valid or not. If it appears to be invalid, we are supposed to assign names to the letters in the proof and check it in a World, but when I do it still checks out.

When I try to prove it in Fitch, I get all the way to the bottom, with (Q^R) as the last step of the subproof and I'm trying to apply -->Intro to join
(NOT)P --> (Q ^ R) but it's not checking out because it says that it needs a support step to be cited.

P --> Q Premise
R ^ S Premise
------------------
(NOT)P --> (Q ^ R) Goal

logic proof language fitch

share|improve this question

edited 1 hour ago

Frank Hubeny

11k51559

asked 2 hours ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Nope, that is not valid. Are you missing a 'not' before that first P?
  
  – Graham Kemp
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is it not valid because there's no (NOT)P in the premises? Since it's not known? No, it's not missing. The first premise is P --> Q.
  
  – fitch-is-killing-me
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Then you cannot conclude ¬P → (Q ˄ R) from those premises, although you may conclude P → (Q ˄ R)
  
  – Graham Kemp
  2 hours ago
  

  
  
  
  

add a comment | 

I'm having trouble with a proof and I'm not sure if it's valid or not. If it appears to be invalid, we are supposed to assign names to the letters in the proof and check it in a World, but when I do it still checks out.

When I try to prove it in Fitch, I get all the way to the bottom, with (Q^R) as the last step of the subproof and I'm trying to apply -->Intro to join
(NOT)P --> (Q ^ R) but it's not checking out because it says that it needs a support step to be cited.

P --> Q Premise
R ^ S Premise
------------------
(NOT)P --> (Q ^ R) Goal

logic proof language fitch

share|improve this question

edited 1 hour ago

Frank Hubeny

11k51559

asked 2 hours ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

P --> Q Premise
R ^ S Premise
------------------
(NOT)P --> (Q ^ R) Goal

share|improve this question

edited 1 hour ago

Frank Hubeny

11k51559

asked 2 hours ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

fitch-is-killing-me 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 question

edited 1 hour ago

Frank Hubeny

11k51559

asked 2 hours ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

edited 1 hour ago

Frank Hubeny

11k51559

edited 1 hour ago

Frank Hubeny

11k51559

asked 2 hours ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

asked 2 hours ago

fitch-is-killing-mefitch-is-killing-me

61

2 Answers
2

active

oldest

votes

2

P → Q, R ˄ S |- ¬P → (Q ˄ R) is not syntactically valid.

An assumption of ¬P will not allow you to eliminate the conditional in P → Q to derive Q .

Is there a typo?

P → Q, R ˄ S |- P → (Q ˄ R) is valid.

 1| P → Q
 2|_ R ˄ S
 3| R ˄E, 2
 4| |_ P
 5| | Q →E, 1,4
 6| | Q ˄ R ˄I, 3,5
 7| P → (Q ˄ R) →I, 4-6

---

For an argument to be semantically valid, the conclusion must be demonstrably true in all interpretations where the premises are -- it is not enough to find just one. A proof is semantically invalid when the exists some interpretation where all the premises are true, but the conclusion is false. One such counter example is enough to disprove an argument.

In the interpretation of P=f,Q=f, R=t, S=t, the premises P → Q, R ˄ S, are both true, but ¬P → (Q ˄ R) is false.

share|improve this answer

edited 1 hour ago

answered 2 hours ago

Graham KempGraham Kemp

1,10919

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No, there's no typo. I imagined that this proof wasn't valid simply because I coulnd't conclude (NOT)P from anywhere. So, when I replace the letters with cube(a) , cube(b), why is it showing, in the world, that the proof is valid? Should I choose names that I know will prove it's invalid?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  

add a comment | 

1

By using a truth table generator one can show that one of the set of valuations for the sentence letters leads to the result being false.

To see this, use this input ((P=>Q)&&(R&&S))=>(~P=>(Q&&R)) in the Stanford Truth Table Tool. When P and Q are false and R and S are true then the conditional with the conjunction of the premises as antecedent and the goal as consequent is false.

This would mean that one could not derive the result.

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That makes sense. So the reason that the proof is invalid, is because out of all the results, there's only 1 that is false? In a valid proof, they should all be true?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That's right. If there were only T values then the statement would be a tautology based on the truth table and a derivation or proof could be found that would be valid.
  
  – Frank Hubeny
  1 hour ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "265"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

fitch-is-killing-me is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphilosophy.stackexchange.com%2fquestions%2f63232%2ffitch-proof-question%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

P → Q, R ˄ S |- ¬P → (Q ˄ R) is not syntactically valid.

An assumption of ¬P will not allow you to eliminate the conditional in P → Q to derive Q .

Is there a typo?

P → Q, R ˄ S |- P → (Q ˄ R) is valid.

 1| P → Q
 2|_ R ˄ S
 3| R ˄E, 2
 4| |_ P
 5| | Q →E, 1,4
 6| | Q ˄ R ˄I, 3,5
 7| P → (Q ˄ R) →I, 4-6

---

For an argument to be semantically valid, the conclusion must be demonstrably true in all interpretations where the premises are -- it is not enough to find just one. A proof is semantically invalid when the exists some interpretation where all the premises are true, but the conclusion is false. One such counter example is enough to disprove an argument.

In the interpretation of P=f,Q=f, R=t, S=t, the premises P → Q, R ˄ S, are both true, but ¬P → (Q ˄ R) is false.

share|improve this answer

edited 1 hour ago

answered 2 hours ago

Graham KempGraham Kemp

1,10919

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No, there's no typo. I imagined that this proof wasn't valid simply because I coulnd't conclude (NOT)P from anywhere. So, when I replace the letters with cube(a) , cube(b), why is it showing, in the world, that the proof is valid? Should I choose names that I know will prove it's invalid?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  

add a comment | 

2

P → Q, R ˄ S |- ¬P → (Q ˄ R) is not syntactically valid.

An assumption of ¬P will not allow you to eliminate the conditional in P → Q to derive Q .

Is there a typo?

P → Q, R ˄ S |- P → (Q ˄ R) is valid.

 1| P → Q
 2|_ R ˄ S
 3| R ˄E, 2
 4| |_ P
 5| | Q →E, 1,4
 6| | Q ˄ R ˄I, 3,5
 7| P → (Q ˄ R) →I, 4-6

---

For an argument to be semantically valid, the conclusion must be demonstrably true in all interpretations where the premises are -- it is not enough to find just one. A proof is semantically invalid when the exists some interpretation where all the premises are true, but the conclusion is false. One such counter example is enough to disprove an argument.

In the interpretation of P=f,Q=f, R=t, S=t, the premises P → Q, R ˄ S, are both true, but ¬P → (Q ˄ R) is false.

share|improve this answer

edited 1 hour ago

answered 2 hours ago

Graham KempGraham Kemp

1,10919

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No, there's no typo. I imagined that this proof wasn't valid simply because I coulnd't conclude (NOT)P from anywhere. So, when I replace the letters with cube(a) , cube(b), why is it showing, in the world, that the proof is valid? Should I choose names that I know will prove it's invalid?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  

add a comment | 

P → Q, R ˄ S |- ¬P → (Q ˄ R) is not syntactically valid.

An assumption of ¬P will not allow you to eliminate the conditional in P → Q to derive Q .

Is there a typo?

P → Q, R ˄ S |- P → (Q ˄ R) is valid.

 1| P → Q
 2|_ R ˄ S
 3| R ˄E, 2
 4| |_ P
 5| | Q →E, 1,4
 6| | Q ˄ R ˄I, 3,5
 7| P → (Q ˄ R) →I, 4-6

---

For an argument to be semantically valid, the conclusion must be demonstrably true in all interpretations where the premises are -- it is not enough to find just one. A proof is semantically invalid when the exists some interpretation where all the premises are true, but the conclusion is false. One such counter example is enough to disprove an argument.

In the interpretation of P=f,Q=f, R=t, S=t, the premises P → Q, R ˄ S, are both true, but ¬P → (Q ˄ R) is false.

share|improve this answer

edited 1 hour ago

answered 2 hours ago

Graham KempGraham Kemp

1,10919

 1| P → Q
 2|_ R ˄ S
 3| R ˄E, 2
 4| |_ P
 5| | Q →E, 1,4
 6| | Q ˄ R ˄I, 3,5
 7| P → (Q ˄ R) →I, 4-6

share|improve this answer

edited 1 hour ago

answered 2 hours ago

Graham KempGraham Kemp

1,10919

answered 2 hours ago

Graham KempGraham Kemp

1,10919

answered 2 hours ago

Graham KempGraham Kemp

1,10919

1

By using a truth table generator one can show that one of the set of valuations for the sentence letters leads to the result being false.

To see this, use this input ((P=>Q)&&(R&&S))=>(~P=>(Q&&R)) in the Stanford Truth Table Tool. When P and Q are false and R and S are true then the conditional with the conjunction of the premises as antecedent and the goal as consequent is false.

This would mean that one could not derive the result.

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That makes sense. So the reason that the proof is invalid, is because out of all the results, there's only 1 that is false? In a valid proof, they should all be true?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That's right. If there were only T values then the statement would be a tautology based on the truth table and a derivation or proof could be found that would be valid.
  
  – Frank Hubeny
  1 hour ago
  

  
  
  
  

add a comment | 

1

By using a truth table generator one can show that one of the set of valuations for the sentence letters leads to the result being false.

To see this, use this input ((P=>Q)&&(R&&S))=>(~P=>(Q&&R)) in the Stanford Truth Table Tool. When P and Q are false and R and S are true then the conditional with the conjunction of the premises as antecedent and the goal as consequent is false.

This would mean that one could not derive the result.

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That makes sense. So the reason that the proof is invalid, is because out of all the results, there's only 1 that is false? In a valid proof, they should all be true?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That's right. If there were only T values then the statement would be a tautology based on the truth table and a derivation or proof could be found that would be valid.
  
  – Frank Hubeny
  1 hour ago
  

  
  
  
  

add a comment | 

By using a truth table generator one can show that one of the set of valuations for the sentence letters leads to the result being false.

To see this, use this input ((P=>Q)&&(R&&S))=>(~P=>(Q&&R)) in the Stanford Truth Table Tool. When P and Q are false and R and S are true then the conditional with the conjunction of the premises as antecedent and the goal as consequent is false.

This would mean that one could not derive the result.

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559