Give a categorical proof of p -> [q -> q] The 2019 Stack Overflow Developer Survey Results Are InFalse(ified) AxiomsHilbert calculus: Proof that every provable formula has a proofHelp with axiom in propositional logicWhy does adding material implication as an axiom to propositional calculus make every formula provable?$S5$ proof of $diamond square pto square p$Can someone break this Propositional logic formulae down for me via truth tableProof of $(neg A supset A) supset A$Can't seem to show that (P∧R)⇒¬(Q∧¬(P∧R)) is a tautologyUnderstanding of Material Implication$(p wedge q) wedge p$ convert to CNF
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Give a categorical proof of p -> [q -> q]
The 2019 Stack Overflow Developer Survey Results Are InFalse(ified) AxiomsHilbert calculus: Proof that every provable formula has a proofHelp with axiom in propositional logicWhy does adding material implication as an axiom to propositional calculus make every formula provable?$S5$ proof of $diamond square pto square p$Can someone break this Propositional logic formulae down for me via truth tableProof of $(neg A supset A) supset A$Can't seem to show that (P∧R)⇒¬(Q∧¬(P∧R)) is a tautologyUnderstanding of Material Implication$(p wedge q) wedge p$ convert to CNF
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I am doing Fitch's Exercises of Symbolic Logic, Chapter 1. This is the first exercise. We have so far axioms such as the distributivity axiom, the axiom of conditioned repetition, the transitivity of implication. I really don't know how to show that this is a theorem.
logic propositional-calculus
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$begingroup$
I am doing Fitch's Exercises of Symbolic Logic, Chapter 1. This is the first exercise. We have so far axioms such as the distributivity axiom, the axiom of conditioned repetition, the transitivity of implication. I really don't know how to show that this is a theorem.
logic propositional-calculus
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add a comment |
$begingroup$
I am doing Fitch's Exercises of Symbolic Logic, Chapter 1. This is the first exercise. We have so far axioms such as the distributivity axiom, the axiom of conditioned repetition, the transitivity of implication. I really don't know how to show that this is a theorem.
logic propositional-calculus
$endgroup$
I am doing Fitch's Exercises of Symbolic Logic, Chapter 1. This is the first exercise. We have so far axioms such as the distributivity axiom, the axiom of conditioned repetition, the transitivity of implication. I really don't know how to show that this is a theorem.
logic propositional-calculus
logic propositional-calculus
edited Mar 23 at 4:27
MJD
47.8k29216397
47.8k29216397
asked Mar 23 at 3:59
simulacrasimulacra
463
463
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add a comment |
2 Answers
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thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)
1) q->(q->q) axiom of conditioned repetition
2) q->(q->q)->q axiom of conditioned repetition
3) (q->(q->q))->(q->q) by 2 and distributivity
4) q->q by 1, 3, modus ponies
5) p->(q->q) axiom of conditioned repetition
$endgroup$
$begingroup$
modus ponies? $ddotsmile$
$endgroup$
– Graham Kemp
Mar 23 at 14:01
add a comment |
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Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.
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2 Answers
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2 Answers
2
active
oldest
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active
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$begingroup$
thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)
1) q->(q->q) axiom of conditioned repetition
2) q->(q->q)->q axiom of conditioned repetition
3) (q->(q->q))->(q->q) by 2 and distributivity
4) q->q by 1, 3, modus ponies
5) p->(q->q) axiom of conditioned repetition
$endgroup$
$begingroup$
modus ponies? $ddotsmile$
$endgroup$
– Graham Kemp
Mar 23 at 14:01
add a comment |
$begingroup$
thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)
1) q->(q->q) axiom of conditioned repetition
2) q->(q->q)->q axiom of conditioned repetition
3) (q->(q->q))->(q->q) by 2 and distributivity
4) q->q by 1, 3, modus ponies
5) p->(q->q) axiom of conditioned repetition
$endgroup$
$begingroup$
modus ponies? $ddotsmile$
$endgroup$
– Graham Kemp
Mar 23 at 14:01
add a comment |
$begingroup$
thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)
1) q->(q->q) axiom of conditioned repetition
2) q->(q->q)->q axiom of conditioned repetition
3) (q->(q->q))->(q->q) by 2 and distributivity
4) q->q by 1, 3, modus ponies
5) p->(q->q) axiom of conditioned repetition
$endgroup$
thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)
1) q->(q->q) axiom of conditioned repetition
2) q->(q->q)->q axiom of conditioned repetition
3) (q->(q->q))->(q->q) by 2 and distributivity
4) q->q by 1, 3, modus ponies
5) p->(q->q) axiom of conditioned repetition
answered Mar 23 at 7:09
simulacrasimulacra
463
463
$begingroup$
modus ponies? $ddotsmile$
$endgroup$
– Graham Kemp
Mar 23 at 14:01
add a comment |
$begingroup$
modus ponies? $ddotsmile$
$endgroup$
– Graham Kemp
Mar 23 at 14:01
$begingroup$
modus ponies? $ddotsmile$
$endgroup$
– Graham Kemp
Mar 23 at 14:01
$begingroup$
modus ponies? $ddotsmile$
$endgroup$
– Graham Kemp
Mar 23 at 14:01
add a comment |
$begingroup$
Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.
$endgroup$
add a comment |
$begingroup$
Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.
$endgroup$
add a comment |
$begingroup$
Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.
$endgroup$
Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.
answered Mar 23 at 4:23
community wiki
MJD
add a comment |
add a comment |
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