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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 11 hours ago
user538118user538118
132
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user538118 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 11 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
10 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 11 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 11 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 11 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 11 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 11 hours ago
user538118user538118
132
asked 11 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
10 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
10 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
10 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
10 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 10 hours ago
AdamAdam
874112
Aha that makes sense! Thank you.
– user538118
9 hours ago
Happy it helped :)
– Adam
7 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 1 hour ago
Graham KempGraham Kemp
1,151110
add a comment |
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2 Answers
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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 10 hours ago
AdamAdam
874112
Aha that makes sense! Thank you.
– user538118
9 hours ago
Happy it helped :)
– Adam
7 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 10 hours ago
AdamAdam
874112
Aha that makes sense! Thank you.
– user538118
9 hours ago
Happy it helped :)
– Adam
7 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 10 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 10 hours ago
AdamAdam
874112
edited 9 hours ago
answered 10 hours ago
AdamAdam
874112
answered 10 hours ago
AdamAdam
874112
answered 10 hours ago
AdamAdam
874112
874112
Aha that makes sense! Thank you.
– user538118
9 hours ago
Happy it helped :)
– Adam
7 hours ago
add a comment |
Aha that makes sense! Thank you.
– user538118
9 hours ago
Happy it helped :)
– Adam
7 hours ago
Aha that makes sense! Thank you.
– user538118
9 hours ago
Aha that makes sense! Thank you.
– user538118
9 hours ago
Happy it helped :)
– Adam
7 hours ago
Happy it helped :)
– Adam
7 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Graham KempGraham Kemp
1,151110
answered 1 hour ago
Graham KempGraham Kemp
1,151110
answered 1 hour ago
Graham KempGraham Kemp
1,151110
answered 1 hour 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.
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You have only exchange P and Q... You have derived P ^ ~P from ~Q. Thus, apply RAA rule to get ~~Q.
– Mauro ALLEGRANZA
10 hours ago