TSTP Solution File: SEV299^5 by cocATP---0.2.0
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%------------------------------------------------------------------------------
% 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
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