%------------------------------------------------------------------------------ % File : cocATP---0.2.0 % Problem : SEV388^5 : TPTP v6.1.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p % Computer : n104.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 13:34:08 EDT 2014 % Result : Theorem 0.40s % Output : Proof 0.40s % Verified : % SZS Type : None (Parsing solution fails) % Syntax : Number of formulae : 0 % Comments : %------------------------------------------------------------------------------ %----ERROR: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % % Problem : SEV388^5 : TPTP v6.1.0. Released v4.0.0. % % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p % % Computer : n104.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 09:04:01 CDT 2014 % % CPUTime : 0.40 % 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 0x1b69878>, <kernel.DependentProduct object at 0x1b0b9e0>) of role type named cS % Using role type % Declaring cS:(fofType->Prop) % FOF formula (<kernel.Constant object at 0x1b60128>, <kernel.DependentProduct object at 0x1b0bf80>) of role type named cR % Using role type % Declaring cR:(fofType->Prop) % FOF formula ((((eq (fofType->Prop)) cR) cS)->(forall (Xx:fofType), ((cR Xx)->(cS Xx)))) of role conjecture named cTHM36_pme % Conjecture to prove = ((((eq (fofType->Prop)) cR) cS)->(forall (Xx:fofType), ((cR Xx)->(cS Xx)))):Prop % Parameter fofType_DUMMY:fofType. % We need to prove ['((((eq (fofType->Prop)) cR) cS)->(forall (Xx:fofType), ((cR Xx)->(cS Xx))))'] % Parameter fofType:Type. % Parameter cS:(fofType->Prop). % Parameter cR:(fofType->Prop). % Trying to prove ((((eq (fofType->Prop)) cR) cS)->(forall (Xx:fofType), ((cR Xx)->(cS Xx)))) % Found x0:=(x (fun (x0:(fofType->Prop))=> (x0 Xx))):((cR Xx)->(cS Xx)) % Found (x (fun (x0:(fofType->Prop))=> (x0 Xx))) as proof of ((cR Xx)->(cS Xx)) % Found (fun (Xx:fofType)=> (x (fun (x0:(fofType->Prop))=> (x0 Xx)))) as proof of ((cR Xx)->(cS Xx)) % Found (fun (x:(((eq (fofType->Prop)) cR) cS)) (Xx:fofType)=> (x (fun (x0:(fofType->Prop))=> (x0 Xx)))) as proof of (forall (Xx:fofType), ((cR Xx)->(cS Xx))) % Found (fun (x:(((eq (fofType->Prop)) cR) cS)) (Xx:fofType)=> (x (fun (x0:(fofType->Prop))=> (x0 Xx)))) as proof of ((((eq (fofType->Prop)) cR) cS)->(forall (Xx:fofType), ((cR Xx)->(cS Xx)))) % Got proof (fun (x:(((eq (fofType->Prop)) cR) cS)) (Xx:fofType)=> (x (fun (x0:(fofType->Prop))=> (x0 Xx)))) % Time elapsed = 0.083483s % node=5 cost=-88.000000 depth=3 % :::::::::::::::::::::: % % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p % % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p % (fun (x:(((eq (fofType->Prop)) cR) cS)) (Xx:fofType)=> (x (fun (x0:(fofType->Prop))=> (x0 Xx)))) % % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p % EOF %------------------------------------------------------------------------------