%------------------------------------------------------------------------------ % File : cocATP---0.2.0 % Problem : NUM709^1 : TPTP v7.0.0. Released v3.7.0. % Transfm : none % Format : tptp:raw % Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p % Computer : n154.star.cs.uiowa.edu % Model : x86_64 x86_64 % CPU : Intel(R) Xeon(R) CPU E5-2609 0 2.40GHz % Memory : 32218.625MB % OS : Linux 3.10.0-693.2.2.el7.x86_64 % CPULimit : 300s % DateTime : Mon Jan 8 13:11:32 EST 2018 % Result : Theorem 0.50s % Output : Proof 0.50s % Verified : % SZS Type : None (Parsing solution fails) % Syntax : Number of formulae : 0 % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.00/0.03 % Problem : NUM709^1 : TPTP v7.0.0. Released v3.7.0. % 0.00/0.03 % Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p % 0.02/0.23 % Computer : n154.star.cs.uiowa.edu % 0.02/0.23 % Model : x86_64 x86_64 % 0.02/0.23 % CPU : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz % 0.02/0.23 % Memory : 32218.625MB % 0.02/0.23 % OS : Linux 3.10.0-693.2.2.el7.x86_64 % 0.02/0.23 % CPULimit : 300 % 0.02/0.23 % DateTime : Fri Jan 5 13:10:34 CST 2018 % 0.02/0.23 % CPUTime : % 0.07/0.27 Python 2.7.13 % 0.50/0.90 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox2/benchmark/', '/export/starexec/sandbox2/benchmark/'] % 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59098>, <kernel.Type object at 0x2ae25ae59b48>) of role type named nat_type % 0.50/0.90 Using role type % 0.50/0.90 Declaring nat:Type % 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59710>, <kernel.Constant object at 0x2ae25ae59ea8>) of role type named x % 0.50/0.90 Using role type % 0.50/0.90 Declaring x:nat % 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59bd8>, <kernel.Constant object at 0x2ae25ae59098>) of role type named y % 0.50/0.90 Using role type % 0.50/0.90 Declaring y:nat % 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59710>, <kernel.DependentProduct object at 0x2ae25aa8e0e0>) of role type named pl % 0.50/0.90 Using role type % 0.50/0.90 Declaring pl:(nat->(nat->nat)) % 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59098>, <kernel.DependentProduct object at 0x2ae25aa8ee60>) of role type named ts % 0.50/0.90 Using role type % 0.50/0.90 Declaring ts:(nat->(nat->nat)) % 0.50/0.90 FOF formula (<kernel.Constant object at 0x2ae25ae59710>, <kernel.DependentProduct object at 0x2ae25aa8eb00>) of role type named suc % 0.50/0.90 Using role type % 0.50/0.90 Declaring suc:(nat->nat) % 0.50/0.90 FOF formula (forall (Xx:nat) (Xy:nat), (((eq nat) ((ts (suc Xx)) Xy)) ((pl ((ts Xx) Xy)) Xy))) of role axiom named satz28d % 0.50/0.90 A new axiom: (forall (Xx:nat) (Xy:nat), (((eq nat) ((ts (suc Xx)) Xy)) ((pl ((ts Xx) Xy)) Xy))) % 0.50/0.90 FOF formula (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)) of role conjecture named satz28h % 0.50/0.90 Conjecture to prove = (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)):Prop % 0.50/0.90 We need to prove ['(((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y))'] % 0.50/0.90 Parameter nat:Type. % 0.50/0.90 Parameter x:nat. % 0.50/0.90 Parameter y:nat. % 0.50/0.90 Parameter pl:(nat->(nat->nat)). % 0.50/0.90 Parameter ts:(nat->(nat->nat)). % 0.50/0.90 Parameter suc:(nat->nat). % 0.50/0.90 Axiom satz28d:(forall (Xx:nat) (Xy:nat), (((eq nat) ((ts (suc Xx)) Xy)) ((pl ((ts Xx) Xy)) Xy))). % 0.50/0.90 Trying to prove (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)) % 0.50/0.90 Found eq_ref00:=(eq_ref0 ((ts (suc x)) y)):(((eq nat) ((ts (suc x)) y)) ((ts (suc x)) y)) % 0.50/0.90 Found (eq_ref0 ((ts (suc x)) y)) as proof of (((eq nat) ((ts (suc x)) y)) ((ts (suc x)) y)) % 0.50/0.90 Found ((eq_ref nat) ((ts (suc x)) y)) as proof of (((eq nat) ((ts (suc x)) y)) ((ts (suc x)) y)) % 0.50/0.90 Found ((eq_ref nat) ((ts (suc x)) y)) as proof of (((eq nat) ((ts (suc x)) y)) ((ts (suc x)) y)) % 0.50/0.90 Found (satz28d000 ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)) % 0.50/0.90 Found ((satz28d00 (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)) % 0.50/0.90 Found (((satz28d0 y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)) % 0.50/0.90 Found ((((satz28d x) y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)) % 0.50/0.90 Found ((((satz28d x) y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) as proof of (((eq nat) ((pl ((ts x) y)) y)) ((ts (suc x)) y)) % 0.50/0.90 Got proof ((((satz28d x) y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) % 0.50/0.90 Time elapsed = 0.176284s % 0.50/0.90 node=27 cost=-103.000000 depth=7 % 0.50/0.90:::::::::::::::::::::: % 0.50/0.90 % SZS status Theorem for /export/starexec/sandbox2/benchmark/theBenchmark.p % 0.50/0.90 % SZS output start Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p % 0.50/0.90 ((((satz28d x) y) (fun (x1:nat)=> (((eq nat) x1) ((ts (suc x)) y)))) ((eq_ref nat) ((ts (suc x)) y))) % 0.50/0.90 % SZS output end Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p %------------------------------------------------------------------------------