meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 5 months |
| seen | Nov 13 '12 at 0:13 | |
| stats | profile views | 11 |
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Nov 12 |
answered | Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms |
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Nov 12 |
revised |
Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms deleted 2991 characters in body |
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Nov 12 |
comment |
Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms I think I found a way to avoid using the meta rule $ \varnothing \vdash p $ implies $ \Gamma \vdash p $. I added it to my original posting. Thank you again for your help. |
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Nov 12 |
revised |
Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms added 1094 characters in body |
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Nov 11 |
revised |
Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms improved formatting |
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Nov 11 |
comment |
Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms Thank you for your observations, I have responded more fully to them by adding to my original posting. |
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Nov 11 |
awarded | Editor |
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Nov 11 |
accepted | Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms |
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Nov 11 |
revised |
Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms Long response to Mr. Makholm |
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Nov 10 |
asked | Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms |
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May 17 |
awarded | Nice Question |
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Dec 23 |
awarded | Supporter |
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Dec 23 |
awarded | Scholar |
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Dec 23 |
accepted | Proving $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$ |
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Dec 23 |
comment |
Proving $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$ Thanks to all. I've also learned two new notations from your responses. |
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Dec 11 |
comment |
Proving $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$ Finally, no the deduction theorem is not available in this system. |
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Dec 11 |
comment |
Proving $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$ The axioms are taken from page 147 of Tarski's Introduction to Logic. I must (guiltily) add that there are three other axioms: (p <-> q) -> (p -> q), (p <-> q) -> (q -> p), (p -> q) -> ((q -> p) -> (p <-> q). I couldn't see how these axioms, which basically define equivalence, would be necessary. But following your remarks I suspect the last axiom above in particular might be crucial. Thanks for your help so far! |
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Dec 11 |
awarded | Student |
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Dec 11 |
asked | Proving $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$ |
