If you are looking for a sequent calculus prover for
first-order classical logic (LK),
please check
Sequent Prover (seqprover).
Examples
All the following examples are valid in classical logic. When the prover fails in proving one of these formulas below, it means that this formula is not intuitionistically valid. Then, you can test its validity in G4c in the prover above and compare the ouput with the result provided by Naoyuki Tamura's prover.
p/\top <--> p
p\/bot <--> p
p/\bot <--> bot
p\/top <--> top
p/\p <--> p
p\/p <--> p
p/\q <--> q/\p
p\/q <--> q\/p
p/\q --> p
p/\q --> q
p --> p\/q
q --> p\/q
p/\(q/\r) <--> (p/\q)/\r
p\/(q\/r) <--> (p\/q)\/r
p/\(q\/r) <--> (p/\q)\/(p/\r)
p\/(q/\r) <--> (p\/q)/\(p\/r)
p/\(q\/r) --> (p\/q)/\(p\/r)
(p/\q)\/(p/\r) --> p\/(q/\r)
~ ~ p <--> p
p/\ ~p <--> bot
p\/ ~p <--> top
~(p/\q) <--> ~p\/ ~q
~(p\/q) <--> ~p/\ ~q
p-> q <--> ~p\/q
~p <--> p-> bot
p-> q <--> ~q-> ~p
p-> q-> r <--> q-> p-> r
p-> q-> r <--> (p/\q)-> r
(p-> q)/\(p-> r) <--> p-> q/\r
(p-> r)/\(q-> r) <--> (p\/q)-> r
p-> q, q-> r --> p-> r
p --> (q-> p)
(p-> (q-> r))--> (p-> q)-> (p-> r)
p-> ~q --> (p-> q)-> ~p
~p-> ~q --> (~p-> q)-> p
p/\(p-> q)/\(p/\q-> s)--> s
top --> p\/(p-> q)
((p-> q)-> p)--> p
top-->X@p(X)-> p(a)/\p(b)
p(a)\/p(b) --> X#p(X)
X@X@p(X) <--> X@p(X)
X#X#p(X) <--> X#p(X)
X@(p(X)/\q(X)) <--> X@p(X) /\ X@q(X)
X#(p(X)\/q(X)) <--> X#p(X) \/ X#q(X)
~X@p(X) <--> X#(~p(X))
~X#p(X) <--> X@(~p(X))
~X@(p(X)-> q(X)) <--> X#(p(X)/\ ~q(X))
~X#(p(X)/\q(X)) <--> X@(p(X)-> ~q(X))
SLOW!
X@p(X) --> X#p(X)
top --> Y#X@(p(X)->p(Y))
X#Y@p(X,Y) --> Y@X#p(X,Y)
(p(a)/\p(b)-> q)-> X#(p(X)-> q)
n(0), X@(n(X)-> n(s(X))) --> n(s(s(0)))