What are the conditions for RAA?What are the necessary conditions for an action to be regarded as a free choice?Is “nothing” easy to understand?Succinct argument for the fundamental role of binary digits as information unitsConditions for DeductionProof of a Conditional and Discharged AssumptionsAre all sufficient conditions necessary?When does ℱx ≡ (x)ℱx ? What if x has the same domain in both?Satisfaction/validity and variable assignmentsWe do everything for pleasureWhy did we define vacuous statements as true rather than false?
How to melt snow without fire or body heat?
How to cut a climbing rope?
Is it truly impossible to tell what a CPU is doing?
Why are GND pads often only connected by four traces?
Which European Languages are not Indo-European?
If a (distance) metric on a connected Riemannian manifold locally agrees with the Riemannian metric, is it equal to the induced metric?
Can my floppy disk still work without a shutter spring?
Do photons bend spacetime or not?
Why would a rational buyer offer to buy with no conditions precedent?
Did 20% of US soldiers in Vietnam use heroin, 95% of whom quit afterwards?
Why haven't we yet tried accelerating a space station with people inside to a near light speed?
Manager questioning my time estimates for a project
How do I get the ς (final sigma) symbol?
Time complexity of an algorithm: Is it important to state the base of the logarithm?
Where is Jon going?
Is superuser the same as root?
Function argument returning void or non-void type
Why did Jon Snow do this immoral act if he is so honorable?
How can I make an argument that my time is valuable?
Why is the Eisenstein ideal paper so great?
Finding collisions of the first few bits of a SHA-1 hash
USPS Back Room - Trespassing?
Python program for a simple calculator
The art of clickbait captions
What are the conditions for RAA?
What are the necessary conditions for an action to be regarded as a free choice?Is “nothing” easy to understand?Succinct argument for the fundamental role of binary digits as information unitsConditions for DeductionProof of a Conditional and Discharged AssumptionsAre all sufficient conditions necessary?When does ℱx ≡ (x)ℱx ? What if x has the same domain in both?Satisfaction/validity and variable assignmentsWe do everything for pleasureWhy did we define vacuous statements as true rather than false?
My textbook states that:
In this case, however, what about situations where we can get Q ^ ~Q (sorry, unfamiliar with this formatting) without depending on P? For instance, the proof of EFQ:
1 (1) P A
2 (2) ~P A
3 (3) ~Q A
1,2 (4) P ^ ~P 1^2
1,2 (5) ~~Q 3,4 RAA
1,2 (6) Q 5 DN
Why isn't it a problem that line 5 does not depend on line 3, when it is the negation of line 3 that we are proving? If we take the above definition of RAA to be true, don't we need 3 to be a dependency of P ^ ~P?
logic propositional-logic
asked 10 hours ago
user538118user538118
132
New contributor
user538118 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
My textbook states that:
In this case, however, what about situations where we can get Q ^ ~Q (sorry, unfamiliar with this formatting) without depending on P? For instance, the proof of EFQ:
1 (1) P A
2 (2) ~P A
3 (3) ~Q A
1,2 (4) P ^ ~P 1^2
1,2 (5) ~~Q 3,4 RAA
1,2 (6) Q 5 DN
Why isn't it a problem that line 5 does not depend on line 3, when it is the negation of line 3 that we are proving? If we take the above definition of RAA to be true, don't we need 3 to be a dependency of P ^ ~P?
logic propositional-logic
asked 10 hours ago
user538118user538118
132
New contributor
user538118 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You have only exchange P and Q... You have derived P ^ ~P from ~Q. Thus, apply RAA rule to get ~~Q.
– Mauro ALLEGRANZA
9 hours ago
add a comment |
My textbook states that:
In this case, however, what about situations where we can get Q ^ ~Q (sorry, unfamiliar with this formatting) without depending on P? For instance, the proof of EFQ:
1 (1) P A
2 (2) ~P A
3 (3) ~Q A
1,2 (4) P ^ ~P 1^2
1,2 (5) ~~Q 3,4 RAA
1,2 (6) Q 5 DN
Why isn't it a problem that line 5 does not depend on line 3, when it is the negation of line 3 that we are proving? If we take the above definition of RAA to be true, don't we need 3 to be a dependency of P ^ ~P?
logic propositional-logic
asked 10 hours ago
user538118user538118
132
New contributor
user538118 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
My textbook states that:
In this case, however, what about situations where we can get Q ^ ~Q (sorry, unfamiliar with this formatting) without depending on P? For instance, the proof of EFQ:
1 (1) P A
2 (2) ~P A
3 (3) ~Q A
1,2 (4) P ^ ~P 1^2
1,2 (5) ~~Q 3,4 RAA
1,2 (6) Q 5 DN
Why isn't it a problem that line 5 does not depend on line 3, when it is the negation of line 3 that we are proving? If we take the above definition of RAA to be true, don't we need 3 to be a dependency of P ^ ~P?
logic propositional-logic
logic propositional-logic
asked 10 hours ago
user538118user538118
132
New contributor
user538118 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
user538118user538118
132
New contributor
user538118 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
user538118user538118
132
New contributor
user538118 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
user538118user538118
132
asked 10 hours ago
user538118user538118
132
132
New contributor
user538118 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user538118 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You have only exchange P and Q... You have derived P ^ ~P from ~Q. Thus, apply RAA rule to get ~~Q.
– Mauro ALLEGRANZA
9 hours ago
add a comment |
You have only exchange P and Q... You have derived P ^ ~P from ~Q. Thus, apply RAA rule to get ~~Q.
– Mauro ALLEGRANZA
9 hours ago
You have only exchange P and Q... You have derived P ^ ~P from ~Q. Thus, apply RAA rule to get ~~Q.
– Mauro ALLEGRANZA
9 hours ago
You have only exchange P and Q... You have derived P ^ ~P from ~Q. Thus, apply RAA rule to get ~~Q.
– Mauro ALLEGRANZA
9 hours ago
add a comment |
2 Answers
2
active
oldest
votes
It only seems like a problem if we think the set Γ is consistent, in which case we feel like it's P which "makes" the premises inconsistent, and so must be used in the proof. But if Γ is already inconsistent, then we don't need to use P at all. Think about it in cases:
If Γ is consistent, and Γ, P ⊢ Q ∧ ¬Q, then Γ, P is inconsistent, so that Γ ⊢ ¬P
If, on the other hand, Γ is already inconsistent (as in your example proof), then Γ ⊢ ¬P anyways (since inconsistent sets of statements can prove anything).
answered 9 hours ago
AdamAdam
874112
Aha that makes sense! Thank you.
– user538118
8 hours ago
Happy it helped :)
– Adam
6 hours ago
add a comment |
Why isn't it a problem that line 5 does not depend on line 3, when it is the negation of line 3 that we are proving?
Because the rule of RAA states that whenever you have Γ, ~Q Ⱶ P ^ ~P , then you may infer that Γ Ⱶ ~~Q
Now, as Γ Ⱶ P ^ ~P when Γ is P, ~P, therefore we can add any additional premise, such as ~Q, and obtain the required sequent so as to be able to apply Reduction To Absurdity: Γ, ~Q Ⱶ P ^ ~P therefore Γ Ⱶ ~~Q
Here's the full sequent notation for the proof:
1 (1) P Ⱶ P A
2 (2) ~P Ⱶ ~P A
3 (3) ~Q Ⱶ ~Q A
1,2 (4) P, ~P Ⱶ P ^ ~P 1^2
1,2,3 (X) P, ~P, ~Q Ⱶ P ^ ~P 3,4 (skipped this line)
1,2 (5) P, ~P Ⱶ ~~Q X RAA (discharges assumption of ~Q)
1,2 (6) P, ~P Ⱶ Q 5 DN
answered 32 mins ago
Graham KempGraham Kemp
1,151110
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
);
);
user538118 is a new contributor. Be nice, and check out our Code of Conduct.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphilosophy.stackexchange.com%2fquestions%2f63571%2fwhat-are-the-conditions-for-raa%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It only seems like a problem if we think the set Γ is consistent, in which case we feel like it's P which "makes" the premises inconsistent, and so must be used in the proof. But if Γ is already inconsistent, then we don't need to use P at all. Think about it in cases:
If Γ is consistent, and Γ, P ⊢ Q ∧ ¬Q, then Γ, P is inconsistent, so that Γ ⊢ ¬P
If, on the other hand, Γ is already inconsistent (as in your example proof), then Γ ⊢ ¬P anyways (since inconsistent sets of statements can prove anything).
answered 9 hours ago
AdamAdam
874112
Aha that makes sense! Thank you.
– user538118
8 hours ago
Happy it helped :)
– Adam
6 hours ago
add a comment |
It only seems like a problem if we think the set Γ is consistent, in which case we feel like it's P which "makes" the premises inconsistent, and so must be used in the proof. But if Γ is already inconsistent, then we don't need to use P at all. Think about it in cases:
If Γ is consistent, and Γ, P ⊢ Q ∧ ¬Q, then Γ, P is inconsistent, so that Γ ⊢ ¬P
If, on the other hand, Γ is already inconsistent (as in your example proof), then Γ ⊢ ¬P anyways (since inconsistent sets of statements can prove anything).
answered 9 hours ago
AdamAdam
874112
Aha that makes sense! Thank you.
– user538118
8 hours ago
Happy it helped :)
– Adam
6 hours ago
add a comment |
It only seems like a problem if we think the set Γ is consistent, in which case we feel like it's P which "makes" the premises inconsistent, and so must be used in the proof. But if Γ is already inconsistent, then we don't need to use P at all. Think about it in cases:
If Γ is consistent, and Γ, P ⊢ Q ∧ ¬Q, then Γ, P is inconsistent, so that Γ ⊢ ¬P
If, on the other hand, Γ is already inconsistent (as in your example proof), then Γ ⊢ ¬P anyways (since inconsistent sets of statements can prove anything).
answered 9 hours ago
AdamAdam
874112
It only seems like a problem if we think the set Γ is consistent, in which case we feel like it's P which "makes" the premises inconsistent, and so must be used in the proof. But if Γ is already inconsistent, then we don't need to use P at all. Think about it in cases:
If Γ is consistent, and Γ, P ⊢ Q ∧ ¬Q, then Γ, P is inconsistent, so that Γ ⊢ ¬P
If, on the other hand, Γ is already inconsistent (as in your example proof), then Γ ⊢ ¬P anyways (since inconsistent sets of statements can prove anything).
answered 9 hours ago
AdamAdam
874112
edited 8 hours ago
answered 9 hours ago
AdamAdam
874112
answered 9 hours ago
AdamAdam
874112
answered 9 hours ago
AdamAdam
874112
874112
Aha that makes sense! Thank you.
– user538118
8 hours ago
Happy it helped :)
– Adam
6 hours ago
add a comment |
Aha that makes sense! Thank you.
– user538118
8 hours ago
Happy it helped :)
– Adam
6 hours ago
Aha that makes sense! Thank you.
– user538118
8 hours ago
Aha that makes sense! Thank you.
– user538118
8 hours ago
Happy it helped :)
– Adam
6 hours ago
Happy it helped :)
– Adam
6 hours ago
add a comment |
Why isn't it a problem that line 5 does not depend on line 3, when it is the negation of line 3 that we are proving?
Because the rule of RAA states that whenever you have Γ, ~Q Ⱶ P ^ ~P , then you may infer that Γ Ⱶ ~~Q
Now, as Γ Ⱶ P ^ ~P when Γ is P, ~P, therefore we can add any additional premise, such as ~Q, and obtain the required sequent so as to be able to apply Reduction To Absurdity: Γ, ~Q Ⱶ P ^ ~P therefore Γ Ⱶ ~~Q
Here's the full sequent notation for the proof:
1 (1) P Ⱶ P A
2 (2) ~P Ⱶ ~P A
3 (3) ~Q Ⱶ ~Q A
1,2 (4) P, ~P Ⱶ P ^ ~P 1^2
1,2,3 (X) P, ~P, ~Q Ⱶ P ^ ~P 3,4 (skipped this line)
1,2 (5) P, ~P Ⱶ ~~Q X RAA (discharges assumption of ~Q)
1,2 (6) P, ~P Ⱶ Q 5 DN
answered 32 mins ago
Graham KempGraham Kemp
1,151110
add a comment |
Why isn't it a problem that line 5 does not depend on line 3, when it is the negation of line 3 that we are proving?
Because the rule of RAA states that whenever you have Γ, ~Q Ⱶ P ^ ~P , then you may infer that Γ Ⱶ ~~Q
Now, as Γ Ⱶ P ^ ~P when Γ is P, ~P, therefore we can add any additional premise, such as ~Q, and obtain the required sequent so as to be able to apply Reduction To Absurdity: Γ, ~Q Ⱶ P ^ ~P therefore Γ Ⱶ ~~Q
Here's the full sequent notation for the proof:
1 (1) P Ⱶ P A
2 (2) ~P Ⱶ ~P A
3 (3) ~Q Ⱶ ~Q A
1,2 (4) P, ~P Ⱶ P ^ ~P 1^2
1,2,3 (X) P, ~P, ~Q Ⱶ P ^ ~P 3,4 (skipped this line)
1,2 (5) P, ~P Ⱶ ~~Q X RAA (discharges assumption of ~Q)
1,2 (6) P, ~P Ⱶ Q 5 DN
answered 32 mins ago
Graham KempGraham Kemp
1,151110
add a comment |
Why isn't it a problem that line 5 does not depend on line 3, when it is the negation of line 3 that we are proving?
Because the rule of RAA states that whenever you have Γ, ~Q Ⱶ P ^ ~P , then you may infer that Γ Ⱶ ~~Q
Now, as Γ Ⱶ P ^ ~P when Γ is P, ~P, therefore we can add any additional premise, such as ~Q, and obtain the required sequent so as to be able to apply Reduction To Absurdity: Γ, ~Q Ⱶ P ^ ~P therefore Γ Ⱶ ~~Q
Here's the full sequent notation for the proof:
1 (1) P Ⱶ P A
2 (2) ~P Ⱶ ~P A
3 (3) ~Q Ⱶ ~Q A
1,2 (4) P, ~P Ⱶ P ^ ~P 1^2
1,2,3 (X) P, ~P, ~Q Ⱶ P ^ ~P 3,4 (skipped this line)
1,2 (5) P, ~P Ⱶ ~~Q X RAA (discharges assumption of ~Q)
1,2 (6) P, ~P Ⱶ Q 5 DN
answered 32 mins ago
Graham KempGraham Kemp
1,151110
Why isn't it a problem that line 5 does not depend on line 3, when it is the negation of line 3 that we are proving?
Because the rule of RAA states that whenever you have Γ, ~Q Ⱶ P ^ ~P , then you may infer that Γ Ⱶ ~~Q
Now, as Γ Ⱶ P ^ ~P when Γ is P, ~P, therefore we can add any additional premise, such as ~Q, and obtain the required sequent so as to be able to apply Reduction To Absurdity: Γ, ~Q Ⱶ P ^ ~P therefore Γ Ⱶ ~~Q
Here's the full sequent notation for the proof:
1 (1) P Ⱶ P A
2 (2) ~P Ⱶ ~P A
3 (3) ~Q Ⱶ ~Q A
1,2 (4) P, ~P Ⱶ P ^ ~P 1^2
1,2,3 (X) P, ~P, ~Q Ⱶ P ^ ~P 3,4 (skipped this line)
1,2 (5) P, ~P Ⱶ ~~Q X RAA (discharges assumption of ~Q)
1,2 (6) P, ~P Ⱶ Q 5 DN
answered 32 mins ago
Graham KempGraham Kemp
1,151110
answered 32 mins ago
Graham KempGraham Kemp
1,151110
answered 32 mins ago
Graham KempGraham Kemp
1,151110
answered 32 mins ago
Graham KempGraham Kemp
1,151110
1,151110
add a comment |
add a comment |
user538118 is a new contributor. Be nice, and check out our Code of Conduct.
user538118 is a new contributor. Be nice, and check out our Code of Conduct.
user538118 is a new contributor. Be nice, and check out our Code of Conduct.
user538118 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Philosophy Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphilosophy.stackexchange.com%2fquestions%2f63571%2fwhat-are-the-conditions-for-raa%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You have only exchange P and Q... You have derived P ^ ~P from ~Q. Thus, apply RAA rule to get ~~Q.
– Mauro ALLEGRANZA
9 hours ago