%------------------------------------------------------------------------------ % File : cocATP---0.2.0 % Problem : SEV299^5 : TPTP v6.2.0. Bugfixed v6.2.0. % Transfm : none % Format : tptp:raw % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p % Computer : n087.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 2.6.32-504.8.1.el6.x86_64 % CPULimit : 300s % DateTime : Tue Apr 21 16:51:41 EDT 2015 % Result : Theorem 0.06s % Output : Proof 0.06s % 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 : SEV299^5 : TPTP v6.2.0. Bugfixed v6.2.0. % 0.00/0.03 % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p % 0.03/1.08 % Computer : n087.star.cs.uiowa.edu % 0.03/1.08 % Model : x86_64 x86_64 % 0.03/1.08 % CPU : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz % 0.03/1.08 % Memory : 32286.75MB % 0.03/1.08 % OS : Linux 2.6.32-504.8.1.el6.x86_64 % 0.03/1.08 % CPULimit : 300 % 0.03/1.08 % DateTime : Thu Apr 16 12:21:42 CDT 2015 % 0.03/1.08 % CPUTime : % 0.03/1.09 Python 2.7.5 % 0.06/1.56 Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox/benchmark/', '/export/starexec/sandbox/benchmark/'] % 0.06/1.56 FOF formula (<kernel.Constant object at 0xcd8518>, <kernel.DependentProduct object at 0xcd88c0>) of role type named cNAT_type % 0.06/1.56 Using role type % 0.06/1.56 Declaring cNAT:(((fofType->Prop)->Prop)->Prop) % 0.06/1.56 FOF formula (<kernel.Constant object at 0x11091b8>, <kernel.DependentProduct object at 0xcd8488>) of role type named cSUCC_type % 0.06/1.56 Using role type % 0.06/1.56 Declaring cSUCC:(((fofType->Prop)->Prop)->((fofType->Prop)->Prop)) % 0.06/1.56 FOF formula (<kernel.Constant object at 0xcd8c68>, <kernel.DependentProduct object at 0xcd8518>) of role type named cZERO_type % 0.06/1.56 Using role type % 0.06/1.56 Declaring cZERO:((fofType->Prop)->Prop) % 0.06/1.56 FOF formula (((eq ((fofType->Prop)->Prop)) cZERO) (fun (Xp:(fofType->Prop))=> (((ex fofType) (fun (Xx:fofType)=> (Xp Xx)))->False))) of role definition named cZERO_def % 0.06/1.56 A new definition: (((eq ((fofType->Prop)->Prop)) cZERO) (fun (Xp:(fofType->Prop))=> (((ex fofType) (fun (Xx:fofType)=> (Xp Xx)))->False))) % 0.06/1.56 Defined: cZERO:=(fun (Xp:(fofType->Prop))=> (((ex fofType) (fun (Xx:fofType)=> (Xp Xx)))->False)) % 0.06/1.56 FOF formula (((eq (((fofType->Prop)->Prop)->((fofType->Prop)->Prop))) cSUCC) (fun (Xn:((fofType->Prop)->Prop)) (Xp:(fofType->Prop))=> ((ex fofType) (fun (Xx:fofType)=> ((and (Xp Xx)) (Xn (fun (Xt:fofType)=> ((and (not (((eq fofType) Xt) Xx))) (Xp Xt))))))))) of role definition named cSUCC_def % 0.06/1.56 A new definition: (((eq (((fofType->Prop)->Prop)->((fofType->Prop)->Prop))) cSUCC) (fun (Xn:((fofType->Prop)->Prop)) (Xp:(fofType->Prop))=> ((ex fofType) (fun (Xx:fofType)=> ((and (Xp Xx)) (Xn (fun (Xt:fofType)=> ((and (not (((eq fofType) Xt) Xx))) (Xp Xt))))))))) % 0.06/1.56 Defined: cSUCC:=(fun (Xn:((fofType->Prop)->Prop)) (Xp:(fofType->Prop))=> ((ex fofType) (fun (Xx:fofType)=> ((and (Xp Xx)) (Xn (fun (Xt:fofType)=> ((and (not (((eq fofType) Xt) Xx))) (Xp Xt)))))))) % 0.06/1.56 FOF formula (((eq (((fofType->Prop)->Prop)->Prop)) cNAT) (fun (Xn:((fofType->Prop)->Prop))=> (forall (Xp:(((fofType->Prop)->Prop)->Prop)), (((and (Xp cZERO)) (forall (Xx:((fofType->Prop)->Prop)), ((Xp Xx)->(Xp (cSUCC Xx)))))->(Xp Xn))))) of role definition named cNAT_def % 0.06/1.56 A new definition: (((eq (((fofType->Prop)->Prop)->Prop)) cNAT) (fun (Xn:((fofType->Prop)->Prop))=> (forall (Xp:(((fofType->Prop)->Prop)->Prop)), (((and (Xp cZERO)) (forall (Xx:((fofType->Prop)->Prop)), ((Xp Xx)->(Xp (cSUCC Xx)))))->(Xp Xn))))) % 0.06/1.56 Defined: cNAT:=(fun (Xn:((fofType->Prop)->Prop))=> (forall (Xp:(((fofType->Prop)->Prop)->Prop)), (((and (Xp cZERO)) (forall (Xx:((fofType->Prop)->Prop)), ((Xp Xx)->(Xp (cSUCC Xx)))))->(Xp Xn)))) % 0.06/1.56 FOF formula (forall (P:(((fofType->Prop)->Prop)->Prop)), (((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))->(forall (M:((fofType->Prop)->Prop)), ((cNAT M)->(P M))))) of role conjecture named cINDUCTION % 0.06/1.56 Conjecture to prove = (forall (P:(((fofType->Prop)->Prop)->Prop)), (((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))->(forall (M:((fofType->Prop)->Prop)), ((cNAT M)->(P M))))):Prop % 0.06/1.56 Parameter fofType_DUMMY:fofType. % 0.06/1.56 We need to prove ['(forall (P:(((fofType->Prop)->Prop)->Prop)), (((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))->(forall (M:((fofType->Prop)->Prop)), ((cNAT M)->(P M)))))'] % 0.06/1.56 Parameter fofType:Type. % 0.06/1.56 Definition cNAT:=(fun (Xn:((fofType->Prop)->Prop))=> (forall (Xp:(((fofType->Prop)->Prop)->Prop)), (((and (Xp cZERO)) (forall (Xx:((fofType->Prop)->Prop)), ((Xp Xx)->(Xp (cSUCC Xx)))))->(Xp Xn)))):(((fofType->Prop)->Prop)->Prop). % 0.06/1.56 Definition cSUCC:=(fun (Xn:((fofType->Prop)->Prop)) (Xp:(fofType->Prop))=> ((ex fofType) (fun (Xx:fofType)=> ((and (Xp Xx)) (Xn (fun (Xt:fofType)=> ((and (not (((eq fofType) Xt) Xx))) (Xp Xt)))))))):(((fofType->Prop)->Prop)->((fofType->Prop)->Prop)). % 0.06/1.56 Definition cZERO:=(fun (Xp:(fofType->Prop))=> (((ex fofType) (fun (Xx:fofType)=> (Xp Xx)))->False)):((fofType->Prop)->Prop). % 0.06/1.56 Trying to prove (forall (P:(((fofType->Prop)->Prop)->Prop)), (((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))->(forall (M:((fofType->Prop)->Prop)), ((cNAT M)->(P M))))) % 0.06/1.56 Found x:((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X))))) % 0.06/1.57 Found x as proof of ((and (P cZERO)) (forall (Xx:((fofType->Prop)->Prop)), ((P Xx)->(P (cSUCC Xx))))) % 0.06/1.57 Found (x00 x) as proof of (P M) % 0.06/1.57 Found ((x0 P) x) as proof of (P M) % 0.06/1.57 Found (fun (x0:(cNAT M))=> ((x0 P) x)) as proof of (P M) % 0.06/1.57 Found (fun (M:((fofType->Prop)->Prop)) (x0:(cNAT M))=> ((x0 P) x)) as proof of ((cNAT M)->(P M)) % 0.06/1.57 Found (fun (x:((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))) (M:((fofType->Prop)->Prop)) (x0:(cNAT M))=> ((x0 P) x)) as proof of (forall (M:((fofType->Prop)->Prop)), ((cNAT M)->(P M))) % 0.06/1.57 Found (fun (P:(((fofType->Prop)->Prop)->Prop)) (x:((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))) (M:((fofType->Prop)->Prop)) (x0:(cNAT M))=> ((x0 P) x)) as proof of (((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))->(forall (M:((fofType->Prop)->Prop)), ((cNAT M)->(P M)))) % 0.06/1.57 Found (fun (P:(((fofType->Prop)->Prop)->Prop)) (x:((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))) (M:((fofType->Prop)->Prop)) (x0:(cNAT M))=> ((x0 P) x)) as proof of (forall (P:(((fofType->Prop)->Prop)->Prop)), (((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))->(forall (M:((fofType->Prop)->Prop)), ((cNAT M)->(P M))))) % 0.06/1.57 Got proof (fun (P:(((fofType->Prop)->Prop)->Prop)) (x:((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))) (M:((fofType->Prop)->Prop)) (x0:(cNAT M))=> ((x0 P) x)) % 0.06/1.57 Time elapsed = 0.117805s % 0.06/1.57 node=9 cost=51.000000 depth=7 % 0.06/1.57:::::::::::::::::::::: % 0.06/1.57 % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p % 0.06/1.57 % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p % 0.06/1.57 (fun (P:(((fofType->Prop)->Prop)->Prop)) (x:((and (P cZERO)) (forall (X:((fofType->Prop)->Prop)), ((P X)->(P (cSUCC X)))))) (M:((fofType->Prop)->Prop)) (x0:(cNAT M))=> ((x0 P) x)) % 0.06/1.57 % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p %------------------------------------------------------------------------------