%------------------------------------------------------------------------------ % File : cocATP---0.2.0 % Problem : NUM639^1 : TPTP v7.0.0. Released v3.7.0. % Transfm : none % Format : tptp:raw % Command : python CASC.py /export/starexec/sandbox2/benchmark/theBenchmark.p % Computer : n185.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 13:11:15 EST 2018 % Result : Theorem 0.47s % Output : Proof 0.47s % 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 : NUM639^1 : 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 : n185.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 11:19:00 CST 2018 % 0.02/0.23 % CPUTime : % 0.06/0.26 Python 2.7.13 % 0.47/0.85 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox2/benchmark/', '/export/starexec/sandbox2/benchmark/'] % 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba21e7f88c0>, <kernel.Type object at 0x2ba21e8d9758>) of role type named nat_type % 0.47/0.85 Using role type % 0.47/0.85 Declaring nat:Type % 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba21e8d56c8>, <kernel.Constant object at 0x2ba21e8d96c8>) of role type named x % 0.47/0.85 Using role type % 0.47/0.85 Declaring x:nat % 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba21e7f88c0>, <kernel.DependentProduct object at 0x2ba21e8d97a0>) of role type named suc % 0.47/0.85 Using role type % 0.47/0.85 Declaring suc:(nat->nat) % 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba21e7f88c0>, <kernel.DependentProduct object at 0x2ba21e8d9488>) of role type named pl % 0.47/0.85 Using role type % 0.47/0.85 Declaring pl:(nat->(nat->nat)) % 0.47/0.85 FOF formula (<kernel.Constant object at 0x2ba216f5c2d8>, <kernel.Constant object at 0x2ba21e8d9a28>) of role type named n_1 % 0.47/0.85 Using role type % 0.47/0.85 Declaring n_1:nat % 0.47/0.85 FOF formula (forall (Xx:nat), (((eq nat) ((pl Xx) n_1)) (suc Xx))) of role axiom named satz4a % 0.47/0.85 A new axiom: (forall (Xx:nat), (((eq nat) ((pl Xx) n_1)) (suc Xx))) % 0.47/0.85 FOF formula (((eq nat) (suc x)) ((pl x) n_1)) of role conjecture named satz4e % 0.47/0.85 Conjecture to prove = (((eq nat) (suc x)) ((pl x) n_1)):Prop % 0.47/0.85 We need to prove ['(((eq nat) (suc x)) ((pl x) n_1))'] % 0.47/0.85 Parameter nat:Type. % 0.47/0.85 Parameter x:nat. % 0.47/0.85 Parameter suc:(nat->nat). % 0.47/0.85 Parameter pl:(nat->(nat->nat)). % 0.47/0.85 Parameter n_1:nat. % 0.47/0.85 Axiom satz4a:(forall (Xx:nat), (((eq nat) ((pl Xx) n_1)) (suc Xx))). % 0.47/0.85 Trying to prove (((eq nat) (suc x)) ((pl x) n_1)) % 0.47/0.85 Found eq_ref00:=(eq_ref0 ((pl x) n_1)):(((eq nat) ((pl x) n_1)) ((pl x) n_1)) % 0.47/0.85 Found (eq_ref0 ((pl x) n_1)) as proof of (((eq nat) ((pl x) n_1)) ((pl x) n_1)) % 0.47/0.85 Found ((eq_ref nat) ((pl x) n_1)) as proof of (((eq nat) ((pl x) n_1)) ((pl x) n_1)) % 0.47/0.85 Found ((eq_ref nat) ((pl x) n_1)) as proof of (((eq nat) ((pl x) n_1)) ((pl x) n_1)) % 0.47/0.85 Found (satz4a00 ((eq_ref nat) ((pl x) n_1))) as proof of (((eq nat) (suc x)) ((pl x) n_1)) % 0.47/0.85 Found ((satz4a0 (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1))) as proof of (((eq nat) (suc x)) ((pl x) n_1)) % 0.47/0.85 Found (((satz4a x) (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1))) as proof of (((eq nat) (suc x)) ((pl x) n_1)) % 0.47/0.85 Found (((satz4a x) (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1))) as proof of (((eq nat) (suc x)) ((pl x) n_1)) % 0.47/0.85 Got proof (((satz4a x) (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1))) % 0.47/0.85 Time elapsed = 0.147780s % 0.47/0.85 node=25 cost=-130.000000 depth=6 % 0.47/0.85:::::::::::::::::::::: % 0.47/0.85 % SZS status Theorem for /export/starexec/sandbox2/benchmark/theBenchmark.p % 0.47/0.85 % SZS output start Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p % 0.47/0.85 (((satz4a x) (fun (x1:nat)=> (((eq nat) x1) ((pl x) n_1)))) ((eq_ref nat) ((pl x) n_1))) % 0.47/0.85 % SZS output end Proof for /export/starexec/sandbox2/benchmark/theBenchmark.p %------------------------------------------------------------------------------