%------------------------------------------------------------------------------
% File     : cocATP---0.2.0
% Problem  : SEV400^5 : TPTP v6.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p

% Computer : n180.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 13:34:09 EDT 2014

% Result   : Unknown 0.54s
% Output   : None 
% Verified : 
% SZS Type : None (Parsing solution fails)
% Syntax   : Number of formulae    : 0

% Comments : 
%------------------------------------------------------------------------------
%----NO SOLUTION OUTPUT BY SYSTEM
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% % Problem  : SEV400^5 : TPTP v6.1.0. Released v4.0.0.
% % Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% % Computer : n180.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 09:07:11 CDT 2014
% % CPUTime  : 0.54 
% 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 0xe800e0>, <kernel.DependentProduct object at 0x10dc320>) of role type named cQ
% Using role type
% Declaring cQ:(fofType->Prop)
% FOF formula (<kernel.Constant object at 0xcbd638>, <kernel.DependentProduct object at 0x10dcd88>) of role type named cP
% Using role type
% Declaring cP:(fofType->Prop)
% FOF formula ((ex (fofType->Prop)) (fun (S:(fofType->Prop))=> ((and ((and (forall (Xx:fofType), ((S Xx)->(cP Xx)))) (forall (Xx:fofType), ((S Xx)->(cQ Xx))))) (forall (R:(fofType->Prop)), (((and (forall (Xx:fofType), ((R Xx)->(cP Xx)))) (forall (Xx:fofType), ((R Xx)->(cQ Xx))))->(forall (Xx:fofType), ((R Xx)->(S Xx)))))))) of role conjecture named cTHM590_pme
% Conjecture to prove = ((ex (fofType->Prop)) (fun (S:(fofType->Prop))=> ((and ((and (forall (Xx:fofType), ((S Xx)->(cP Xx)))) (forall (Xx:fofType), ((S Xx)->(cQ Xx))))) (forall (R:(fofType->Prop)), (((and (forall (Xx:fofType), ((R Xx)->(cP Xx)))) (forall (Xx:fofType), ((R Xx)->(cQ Xx))))->(forall (Xx:fofType), ((R Xx)->(S Xx)))))))):Prop
% Parameter fofType_DUMMY:fofType.
% We need to prove ['((ex (fofType->Prop)) (fun (S:(fofType->Prop))=> ((and ((and (forall (Xx:fofType), ((S Xx)->(cP Xx)))) (forall (Xx:fofType), ((S Xx)->(cQ Xx))))) (forall (R:(fofType->Prop)), (((and (forall (Xx:fofType), ((R Xx)->(cP Xx)))) (forall (Xx:fofType), ((R Xx)->(cQ Xx))))->(forall (Xx:fofType), ((R Xx)->(S Xx))))))))']
% Parameter fofType:Type.
% Parameter cQ:(fofType->Prop).
% Parameter cP:(fofType->Prop).
% Trying to prove ((ex (fofType->Prop)) (fun (S:(fofType->Prop))=> ((and ((and (forall (Xx:fofType), ((S Xx)->(cP Xx)))) (forall (Xx:fofType), ((S Xx)->(cQ Xx))))) (forall (R:(fofType->Prop)), (((and (forall (Xx:fofType), ((R Xx)->(cP Xx)))) (forall (Xx:fofType), ((R Xx)->(cQ Xx))))->(forall (Xx:fofType), ((R Xx)->(S Xx))))))))
% Found x00:(R Xx)
% Instantiate: x:=R:(fofType->Prop)
% Found (fun (x00:(R Xx))=> x00) as proof of (x Xx)
% Found (fun (Xx:fofType) (x00:(R Xx))=> x00) as proof of ((R Xx)->(x Xx))
% Found (fun (x0:((and (forall (Xx:fofType), ((R Xx)->(cP Xx)))) (forall (Xx:fofType), ((R Xx)->(cQ Xx))))) (Xx:fofType) (x00:(R Xx))=> x00) as proof of (forall (Xx:fofType), ((R Xx)->(x Xx)))
% Found x0:(x Xx)
% Found x0 as proof of (cP Xx)
% Found (fun (x0:(x Xx))=> x0) as proof of (cP Xx)
% Found (fun (Xx:fofType) (x0:(x Xx))=> x0) as proof of ((x Xx)->(cP Xx))
% Found (fun (Xx:fofType) (x0:(x Xx))=> x0) as proof of (forall (Xx:fofType), ((x Xx)->(cP Xx)))
% Found x0:(x Xx)
% Found x0 as proof of (cQ Xx)
% Found (fun (x0:(x Xx))=> x0) as proof of (cQ Xx)
% Found (fun (Xx:fofType) (x0:(x Xx))=> x0) as proof of ((x Xx)->(cQ Xx))
% Found (fun (Xx:fofType) (x0:(x Xx))=> x0) as proof of (forall (Xx:fofType), ((x Xx)->(cQ Xx)))
% % SZS status GaveUp for /export/starexec/sandbox/benchmark/theBenchmark.p
% EOF
%------------------------------------------------------------------------------