Comments on: This might or might not qualify as a theorem prover

By: sandra

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By: ChristianS

ChristianS — Wed, 13 Jun 2007 18:23:48 +0000

“(==True)” is the same as “id”,
“all (==True)” is the same as “and”,
“all (==False)” is “not . any”.

“consistent” seems wrong with the possibility to give two different arities.

By: syntaxfree (on the run)

syntaxfree (on the run) — Wed, 13 Jun 2007 14:43:35 +0000

Oops. Modus tollens:

*Main> taut2 $ \fire oxygen -> (( fire:-> oxygen) :* (Not oxygen)) :-> (Not fire)
True

Affirming the consequent:

*Main> taut2 $ \f o -> (( f:-> o) :* o) :-> f
False

Denying the antecedent:

*Main> taut2 $ \f o -> (( f:-> o) :* o) :-> f
False

False syllogism:

*Main> taut3 $ \socrates cat mortal -> ((cat :-> mortal) :* (socrates :-> mortal)) :-> socrates :-> cat
False

By: syntaxfree (on the run)

syntaxfree (on the run) — Wed, 13 Jun 2007 14:34:56 +0000

Modus ponens:

*Main> taut2 $ \a b -> ((a :-> b) :* a) :-> b
True

Modus tollens:

*Main> taut2 $ \a b -> ((a :-> b) :* a) :-> b
True

By: syntaxfree (on the run)

syntaxfree (on the run) — Wed, 13 Jun 2007 13:49:20 +0000

Well, yeah. This is the “Mortality of Socrates” theorem:

*Main> taut3 (\s h m -> ((h :-> m) :* (s :-> h)) :-> (s :-> m))
True

Yay my first theorem prover!

By: augustss

augustss — Wed, 13 Jun 2007 13:31:51 +0000

BTW, you can write an overloaded taut that works for any number of arguments.

By: augustss

augustss — Wed, 13 Jun 2007 11:55:54 +0000

It’s some kind of theorem prover.

Do you have some future expansion in mind that made you parametrize Formula?

BTW, a simplification:

truth (a :-> b) = not (truth a) || truth b