%------------------------------------------------------------------------------ % File : cocATP---0.2.0 % Problem : SEV187^5 : TPTP v6.1.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p % Computer : n183.star.cs.uiowa.edu % Model : x86_64 x86_64 % CPU : Intel(R) Xeon(R) CPU E5-2609 0 2.40GHz % Memory : 32286.75MB % OS : Linux 2.6.32-431.20.3.el6.x86_64 % CPULimit : 300s % DateTime : Thu Jul 17 13:33:51 EDT 2014 % Result : Theorem 0.35s % Output : Proof 0.35s % Verified : % SZS Type : None (Parsing solution fails) % Syntax : Number of formulae : 0 % Comments : %------------------------------------------------------------------------------ %----ERROR: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % % Problem : SEV187^5 : TPTP v6.1.0. Released v4.0.0. % % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p % % Computer : n183.star.cs.uiowa.edu % % Model : x86_64 x86_64 % % CPU : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz % % Memory : 32286.75MB % % OS : Linux 2.6.32-431.20.3.el6.x86_64 % % CPULimit : 300 % % DateTime : Thu Jul 17 08:24:46 CDT 2014 % % CPUTime : 0.35 % Python 2.7.5 % Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox/benchmark/', '/export/starexec/sandbox/benchmark/'] % FOF formula (<kernel.Constant object at 0x1cdd758>, <kernel.Type object at 0x1cddc20>) of role type named a_type % Using role type % Declaring a:Type % FOF formula (<kernel.Constant object at 0x1cdb1b8>, <kernel.Type object at 0x1cdd4d0>) of role type named b_type % Using role type % Declaring b:Type % FOF formula (<kernel.Constant object at 0x1cdd5a8>, <kernel.Type object at 0x1cdd2d8>) of role type named c_type % Using role type % Declaring c:Type % FOF formula (forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True)) of role conjecture named cCS3_ALL_THM_pme % Conjecture to prove = (forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True)):Prop % Parameter a_DUMMY:a. % Parameter b_DUMMY:b. % Parameter c_DUMMY:c. % We need to prove ['(forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True))'] % Parameter a:Type. % Parameter b:Type. % Parameter c:Type. % Trying to prove (forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True)) % Found I:True % Found (fun (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I) as proof of True % Found (fun (S:((a->(b->(c->Prop)))->Prop)) (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I) as proof of ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True) % Found (fun (S:((a->(b->(c->Prop)))->Prop)) (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I) as proof of (forall (S:((a->(b->(c->Prop)))->Prop)), ((forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True))->True)) % Got proof (fun (S:((a->(b->(c->Prop)))->Prop)) (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I) % Time elapsed = 0.038902s % node=2 cost=5.000000 depth=2 % :::::::::::::::::::::: % % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p % % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p % (fun (S:((a->(b->(c->Prop)))->Prop)) (x:(forall (Xx:(a->(b->(c->Prop)))), ((S Xx)->True)))=> I) % % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p % EOF %------------------------------------------------------------------------------