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Jan
29
awarded  Popular Question
Nov
12
answered Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms
Nov
12
revised Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms
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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.
Nov
12
revised 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
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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.
Nov
11
awarded  Editor
Nov
11
accepted Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms
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
Nov
10
asked Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms
May
17
awarded  Nice Question
Dec
23
awarded  Supporter
Dec
23
awarded  Scholar
Dec
23
accepted Proving $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$
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.
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.
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!
Dec
11
awarded  Student
Dec
11
asked Proving $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$