exists y. (forall z. P(z,y) ==> P(z,f(x))) /\ P(x,y)>>;;
END_INTERACTIVE;;
(* ------------------------------------------------------------------------- *)
(* More standard tableau procedure, effectively doing DNF incrementally. *)
(* ------------------------------------------------------------------------- *)
let rec tableau (fms,lits,n) cont (env,k) =
if n < 0 then failwith "no proof at this level" else
match fms with
[] -> failwith "tableau: no proof"
| And(p,q)::unexp ->
tableau (p::q::unexp,lits,n) cont (env,k)
| Or(p,q)::unexp ->
tableau (p::unexp,lits,n) (tableau (q::unexp,lits,n) cont) (env,k)
| Forall(x,p)::unexp ->
let y = Var("_" ^ string_of_int k) in
let p' = subst (x |=> y) p in
tableau (p'::unexp@[Forall(x,p)],lits,n-1) cont (env,k+1)
| fm::unexp ->
try tryfind (fun l -> cont(unify_complements env (fm,l),k)) lits
with Failure _ -> tableau (unexp,fm::lits,n) cont (env,k);;
let rec deepen f n =
try print_string "Searching with depth limit ";
print_int n; print_newline(); f n
with Failure _ -> deepen f (n + 1);;
let tabrefute fms =
deepen (fun n -> tableau (fms,[],n) (fun x -> x) (undefined,0); n) 0;;
let tab fm =
let sfm = askolemize(Not(generalize fm)) in
if sfm = False then 0 else tabrefute [sfm];;
(* ------------------------------------------------------------------------- *)
(* Example. *)
(* ------------------------------------------------------------------------- *)
START_INTERACTIVE;;
let p38 = tab
<<(forall x.
P(a) /\ (P(x) ==> (exists y. P(y) /\ R(x,y))) ==>
(exists z w. P(z) /\ R(x,w) /\ R(w,z))) <=>
(forall x.
(~P(a) \/ P(x) \/ (exists z w. P(z) /\ R(x,w) /\ R(w,z))) /\
(~P(a) \/ ~(exists y. P(y) /\ R(x,y)) \/
(exists z w. P(z) /\ R(x,w) /\ R(w,z))))>>;;
END_INTERACTIVE;;
(* ------------------------------------------------------------------------- *)
(* Try to split up the initial formula first; often a big improvement. *)
(* ------------------------------------------------------------------------- *)
let splittab fm =
map tabrefute (simpdnf(askolemize(Not(generalize fm))));;
(* ------------------------------------------------------------------------- *)
(* Example: the Andrews challenge. *)
(* ------------------------------------------------------------------------- *)
START_INTERACTIVE;;
let p34 = splittab
<<((exists x. forall y. P(x) <=> P(y)) <=>
((exists x. Q(x)) <=> (forall y. Q(y)))) <=>
((exists x. forall y. Q(x) <=> Q(y)) <=>
((exists x. P(x)) <=> (forall y. P(y))))>>;;
(* ------------------------------------------------------------------------- *)
(* Another nice example from EWD 1602. *)
(* ------------------------------------------------------------------------- *)
let ewd1062 = splittab
<<(forall x. x <= x) /\
(forall x y z. x <= y /\ y <= z ==> x <= z) /\
(forall x y. f(x) <= y <=> x <= g(y))
==> (forall x y. x <= y ==> f(x) <= f(y)) /\
(forall x y. x <= y ==> g(x) <= g(y))>>;;
END_INTERACTIVE;;
(* ------------------------------------------------------------------------- *)
(* Do all the equality-free Pelletier problems, and more, as examples. *)
(* ------------------------------------------------------------------------- *)
(***********
let p1 = time splittab
< q <=> ~q ==> ~p>>;;
let p2 = time splittab
<<~ ~p <=> p>>;;
let p3 = time splittab
<<~(p ==> q) ==> q ==> p>>;;
let p4 = time splittab
<<~p ==> q <=> ~q ==> p>>;;
let p5 = time splittab
<<(p \/ q ==> p \/ r) ==> p \/ (q ==> r)>>;;
let p6 = time splittab
<
>;;
let p7 = time splittab
<
>;;
let p8 = time splittab
<<((p ==> q) ==> p) ==> p>>;;
let p9 = time splittab
<<(p \/ q) /\ (~p \/ q) /\ (p \/ ~q) ==> ~(~q \/ ~q)>>;;
let p10 = time splittab
<<(q ==> r) /\ (r ==> p /\ q) /\ (p ==> q /\ r) ==> (p <=> q)>>;;
let p11 = time splittab
<
p>>;;
let p12 = time splittab
<<((p <=> q) <=> r) <=> (p <=> (q <=> r))>>;;
let p13 = time splittab
<
(p \/ q) /\ (p \/ r)>>;;
let p14 = time splittab
<<(p <=> q) <=> (q \/ ~p) /\ (~q \/ p)>>;;
let p15 = time splittab
<
q <=> ~p \/ q>>;;
let p16 = time splittab
<<(p ==> q) \/ (q ==> p)>>;;
let p17 = time splittab
<
r) ==> s <=> (~p \/ q \/ s) /\ (~p \/ ~r \/ s)>>;;
(* ------------------------------------------------------------------------- *)
(* Pelletier problems: monadic predicate logic. *)
(* ------------------------------------------------------------------------- *)
let p18 = time splittab
< P(x)>>;;
let p19 = time splittab
< Q(z)) ==> P(x) ==> Q(x)>>;;
let p20 = time splittab
<<(forall x y. exists z. forall w. P(x) /\ Q(y) ==> R(z) /\ U(w))
==> (exists x y. P(x) /\ Q(y)) ==> (exists z. R(z))>>;;
let p21 = time splittab
<<(exists x. P ==> Q(x)) /\ (exists x. Q(x) ==> P)
==> (exists x. P <=> Q(x))>>;;
let p22 = time splittab
<<(forall x. P <=> Q(x)) ==> (P <=> (forall x. Q(x)))>>;;
let p23 = time splittab
<<(forall x. P \/ Q(x)) <=> P \/ (forall x. Q(x))>>;;
let p24 = time splittab
<<~(exists x. U(x) /\ Q(x)) /\
(forall x. P(x) ==> Q(x) \/ R(x)) /\
~(exists x. P(x) ==> (exists x. Q(x))) /\
(forall x. Q(x) /\ R(x) ==> U(x)) ==>
(exists x. P(x) /\ R(x))>>;;
let p25 = time splittab
<<(exists x. P(x)) /\
(forall x. U(x) ==> ~G(x) /\ R(x)) /\
(forall x. P(x) ==> G(x) /\ U(x)) /\
((forall x. P(x) ==> Q(x)) \/ (exists x. Q(x) /\ P(x)))
==> (exists x. Q(x) /\ P(x))>>;;
let p26 = time splittab
<<((exists x. P(x)) <=> (exists x. Q(x))) /\
(forall x y. P(x) /\ Q(y) ==> (R(x) <=> U(y)))
==> ((forall x. P(x) ==> R(x)) <=> (forall x. Q(x) ==> U(x)))>>;;
let p27 = time splittab
<<(exists x. P(x) /\ ~Q(x)) /\
(forall x. P(x) ==> R(x)) /\
(forall x. U(x) /\ V(x) ==> P(x)) /\
(exists x. R(x) /\ ~Q(x))
==> (forall x. U(x) ==> ~R(x))
==> (forall x. U(x) ==> ~V(x))>>;;
let p28 = time splittab
<<(forall x. P(x) ==> (forall x. Q(x))) /\
((forall x. Q(x) \/ R(x)) ==> (exists x. Q(x) /\ R(x))) /\
((exists x. R(x)) ==> (forall x. L(x) ==> M(x))) ==>
(forall x. P(x) /\ L(x) ==> M(x))>>;;
let p29 = time splittab
<<(exists x. P(x)) /\ (exists x. G(x)) ==>
((forall x. P(x) ==> H(x)) /\ (forall x. G(x) ==> J(x)) <=>
(forall x y. P(x) /\ G(y) ==> H(x) /\ J(y)))>>;;
let p30 = time splittab
<<(forall x. P(x) \/ G(x) ==> ~H(x)) /\
(forall x. (G(x) ==> ~U(x)) ==> P(x) /\ H(x))
==> (forall x. U(x))>>;;
let p31 = time splittab
<<~(exists x. P(x) /\ (G(x) \/ H(x))) /\
(exists x. Q(x) /\ P(x)) /\
(forall x. ~H(x) ==> J(x))
==> (exists x. Q(x) /\ J(x))>>;;
let p32 = time splittab
<<(forall x. P(x) /\ (G(x) \/ H(x)) ==> Q(x)) /\
(forall x. Q(x) /\ H(x) ==> J(x)) /\
(forall x. R(x) ==> H(x))
==> (forall x. P(x) /\ R(x) ==> J(x))>>;;
let p33 = time splittab
<<(forall x. P(a) /\ (P(x) ==> P(b)) ==> P(c)) <=>
(forall x. P(a) ==> P(x) \/ P(c)) /\ (P(a) ==> P(b) ==> P(c))>>;;
let p34 = time splittab
<<((exists x. forall y. P(x) <=> P(y)) <=>
((exists x. Q(x)) <=> (forall y. Q(y)))) <=>
((exists x. forall y. Q(x) <=> Q(y)) <=>
((exists x. P(x)) <=> (forall y. P(y))))>>;;
let p35 = time splittab
< (forall x y. P(x,y))>>;;
(* ------------------------------------------------------------------------- *)
(* Full predicate logic (without identity and functions). *)
(* ------------------------------------------------------------------------- *)
let p36 = time splittab
<<(forall x. exists y. P(x,y)) /\
(forall x. exists y. G(x,y)) /\
(forall x y. P(x,y) \/ G(x,y)
==> (forall z. P(y,z) \/ G(y,z) ==> H(x,z)))
==> (forall x. exists y. H(x,y))>>;;
let p37 = time splittab
<<(forall z.
exists w. forall x. exists y. (P(x,z) ==> P(y,w)) /\ P(y,z) /\
(P(y,w) ==> (exists u. Q(u,w)))) /\
(forall x z. ~P(x,z) ==> (exists y. Q(y,z))) /\
((exists x y. Q(x,y)) ==> (forall x. R(x,x))) ==>
(forall x. exists y. R(x,y))>>;;
let p38 = time splittab
<<(forall x.
P(a) /\ (P(x) ==> (exists y. P(y) /\ R(x,y))) ==>
(exists z w. P(z) /\ R(x,w) /\ R(w,z))) <=>
(forall x.
(~P(a) \/ P(x) \/ (exists z w. P(z) /\ R(x,w) /\ R(w,z))) /\
(~P(a) \/ ~(exists y. P(y) /\ R(x,y)) \/
(exists z w. P(z) /\ R(x,w) /\ R(w,z))))>>;;
let p39 = time splittab
<<~(exists x. forall y. P(y,x) <=> ~P(y,y))>>;;
let p40 = time splittab
<<(exists y. forall x. P(x,y) <=> P(x,x))
==> ~(forall x. exists y. forall z. P(z,y) <=> ~P(z,x))>>;;
let p41 = time splittab
<<(forall z. exists y. forall x. P(x,y) <=> P(x,z) /\ ~P(x,x))
==> ~(exists z. forall x. P(x,z))>>;;
let p42 = time splittab
<<~(exists y. forall x. P(x,y) <=> ~(exists z. P(x,z) /\ P(z,x)))>>;;
let p43 = time splittab
<<(forall x y. Q(x,y) <=> forall z. P(z,x) <=> P(z,y))
==> forall x y. Q(x,y) <=> Q(y,x)>>;;
let p44 = time splittab
<<(forall x. P(x) ==> (exists y. G(y) /\ H(x,y)) /\
(exists y. G(y) /\ ~H(x,y))) /\
(exists x. J(x) /\ (forall y. G(y) ==> H(x,y))) ==>
(exists x. J(x) /\ ~P(x))>>;;
let p45 = time splittab
<<(forall x.
P(x) /\ (forall y. G(y) /\ H(x,y) ==> J(x,y)) ==>
(forall y. G(y) /\ H(x,y) ==> R(y))) /\
~(exists y. L(y) /\ R(y)) /\
(exists x. P(x) /\ (forall y. H(x,y) ==>
L(y)) /\ (forall y. G(y) /\ H(x,y) ==> J(x,y))) ==>
(exists x. P(x) /\ ~(exists y. G(y) /\ H(x,y)))>>;;
let p46 = time splittab
<<(forall x. P(x) /\ (forall y. P(y) /\ H(y,x) ==> G(y)) ==> G(x)) /\
((exists x. P(x) /\ ~G(x)) ==>
(exists x. P(x) /\ ~G(x) /\
(forall y. P(y) /\ ~G(y) ==> J(x,y)))) /\
(forall x y. P(x) /\ P(y) /\ H(x,y) ==> ~J(y,x)) ==>
(forall x. P(x) ==> G(x))>>;;
(* ------------------------------------------------------------------------- *)
(* Well-known "Agatha" example; cf. Manthey and Bry, CADE-9. *)
(* ------------------------------------------------------------------------- *)
let p55 = time splittab
< hates(x,y) /\ ~richer(x,y)) /\
(forall x. hates(agatha,x) ==> ~hates(charles,x)) /\
(hates(agatha,agatha) /\ hates(agatha,charles)) /\
(forall x. lives(x) /\ ~richer(x,agatha) ==> hates(butler,x)) /\
(forall x. hates(agatha,x) ==> hates(butler,x)) /\
(forall x. ~hates(x,agatha) \/ ~hates(x,butler) \/ ~hates(x,charles))
==> killed(agatha,agatha) /\
~killed(butler,agatha) /\
~killed(charles,agatha)>>;;
let p57 = time splittab
< P(x,z))
==> P(f(a,b),f(a,c))>>;;
(* ------------------------------------------------------------------------- *)
(* See info-hol, circa 1500. *)
(* ------------------------------------------------------------------------- *)
let p58 = time splittab
< ((P(v) \/ R(w)) /\ (R(z) ==> Q(v))))>>;;
let p59 = time splittab
<<(forall x. P(x) <=> ~P(f(x))) ==> (exists x. P(x) /\ ~P(f(x)))>>;;
let p60 = time splittab
<
exists y. (forall z. P(z,y) ==> P(z,f(x))) /\ P(x,y)>>;;
(* ------------------------------------------------------------------------- *)
(* From Gilmore's classic paper. *)
(* ------------------------------------------------------------------------- *)
(***** This is still too hard for us! Amazing...
let gilmore_1 = time splittab
< G(y)) <=> F(x)) /\
((F(y) ==> H(y)) <=> G(x)) /\
(((F(y) ==> G(y)) ==> H(y)) <=> H(x))
==> F(z) /\ G(z) /\ H(z)>>;;
******)
(*** This is not valid, according to Gilmore
let gilmore_2 = time splittab
< F(z,y)) /\ (F(z,y) <=> F(z,z)) /\ (F(x,y) <=> F(y,x))
==> (F(x,y) <=> F(x,z))>>;;
***)
let gilmore_3 = time splittab
< (G(y) ==> H(x))) ==> F(x,x)) /\
((F(z,x) ==> G(x)) ==> H(z)) /\
F(x,y)
==> F(z,z)>>;;
let gilmore_4 = time splittab
< F(y,z) /\ F(z,z)) /\
(F(x,y) /\ G(x,y) ==> G(x,z) /\ G(z,z))>>;;
let gilmore_5 = time splittab
<<(forall x. exists y. F(x,y) \/ F(y,x)) /\
(forall x y. F(y,x) ==> F(y,y))
==> exists z. F(z,z)>>;;
let gilmore_6 = time splittab
< G(v,u) /\ G(u,x))
==> (exists u. forall v. F(u,y) ==> G(v,u) /\ G(u,y)) \/
(forall u v. exists w. G(v,u) \/ H(w,y,u) ==> G(u,w))>>;;
let gilmore_7 = time splittab
<<(forall x. K(x) ==> exists y. L(y) /\ (F(x,y) ==> G(x,y))) /\
(exists z. K(z) /\ forall u. L(u) ==> F(z,u))
==> exists v w. K(v) /\ L(w) /\ G(v,w)>>;;
let gilmore_8 = time splittab
< (G(y) ==> (forall u. exists v. H(u,v,x)))) ==> F(x,x)) /\
((F(z,x) ==> G(x)) ==> (forall u. exists v. H(u,v,z))) /\
F(x,y)
==> F(z,z)>>;;
let gilmore_9 = time splittab
< (forall u. exists v. F(x,u,v) /\ G(z,u) /\ ~H(x,z))
==> (forall u. exists v. F(x,u,v) /\ G(y,u) /\ ~H(x,y))) /\
((forall u. exists v. F(x,u,v) /\ G(y,u) /\ ~H(x,y))
==> ~(forall u. exists v. F(x,u,v) /\ G(z,u) /\ ~H(x,z))
==> (forall u. exists v. F(y,u,v) /\ G(y,u) /\ ~H(y,x)) /\
(forall u. exists v. F(z,u,v) /\ G(y,u) /\ ~H(z,y)))>>;;
(* ------------------------------------------------------------------------- *)
(* Example from Davis-Putnam papers where Gilmore procedure is poor. *)
(* ------------------------------------------------------------------------- *)
let davis_putnam_example = time splittab
< (F(y,z) /\ F(z,z))) /\
((F(x,y) /\ G(x,y)) ==> (G(x,z) /\ G(z,z)))>>;;
*************)