Play with a backwards prover implementing FOL in LK. This prover uses explicit terms for all on the left and exists on the right.
Code is here.
Alternate resource: Check out Larry Paulson's FOL theorem prover linked from this page: http://www.cl.cam.ac.uk/teaching/1011/LogicProof/
Fire up ghci; load FOL/LK.hs
We prove example formula (forall x. P(x)) --> (forall y. P(F(y))) as follows:
m1 = (replayM s1 [impliesR 1, allR 1, allL (toTerm "F(?b)") 1, axiom 1 1])
As in the other backwards provers, the list is a list of rules. We begin with implies on the right (impliesR 1): This turns the sequent with only an implication
|- ((ALL x. P(x)) --> (ALL y. P(F(y))))into one with terms on both sides of the turnstile.
(ALL x. P(x)) |- (ALL y. P(F(y)))
We continue with all on the right (allR 1), which introduces a new variable ?b, that is fresh:
(ALL x. P(x)) |- P(F(?b)), (ALL y. P(F(y)))Note how it also duplicates the for all, a specialized instance (with the fresh varaibale) and the old quantified instance.
Next we apply exists on the right, providing a term as required by the rule:
P(F(?b)) |- P(F(?b)), (ALL y. P(F(y)))
This is followed with all on the left (allL (toTerm "F(?b)") 1), which also requires a term. We choose "F(?b)" .
We complete the proof with the axiom rule (axiom 1 1), which leaves us with: