%------------------------------------------------------------------------------
% File     : cocATP---0.2.0
% Problem  : SEV193^5 : TPTP v6.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p

% Computer : n098.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 13:33:52 EDT 2014

% Result   : Unknown 0.51s
% Output   : None 
% Verified : 
% SZS Type : None (Parsing solution fails)
% Syntax   : Number of formulae    : 0

% Comments : 
%------------------------------------------------------------------------------
%----NO SOLUTION OUTPUT BY SYSTEM
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% % Problem  : SEV193^5 : TPTP v6.1.0. Released v4.0.0.
% % Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% % Computer : n098.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 08:26:11 CDT 2014
% % CPUTime  : 0.51 
% 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 0x12183b0>, <kernel.DependentProduct object at 0x1218638>) of role type named cT
% Using role type
% Declaring cT:((fofType->Prop)->Prop)
% FOF formula ((forall (S:((fofType->Prop)->Prop)), ((forall (Xx:(fofType->Prop)), ((S Xx)->(cT Xx)))->(cT (fun (Xx:fofType)=> ((ex (fofType->Prop)) (fun (S0:(fofType->Prop))=> ((and (S S0)) (S0 Xx))))))))->(forall (U:(fofType->Prop)) (V:(fofType->Prop)), (((and (cT U)) (cT V))->(cT (fun (Xz:fofType)=> ((or (U Xz)) (V Xz))))))) of role conjecture named cTHM501_pme
% Conjecture to prove = ((forall (S:((fofType->Prop)->Prop)), ((forall (Xx:(fofType->Prop)), ((S Xx)->(cT Xx)))->(cT (fun (Xx:fofType)=> ((ex (fofType->Prop)) (fun (S0:(fofType->Prop))=> ((and (S S0)) (S0 Xx))))))))->(forall (U:(fofType->Prop)) (V:(fofType->Prop)), (((and (cT U)) (cT V))->(cT (fun (Xz:fofType)=> ((or (U Xz)) (V Xz))))))):Prop
% Parameter fofType_DUMMY:fofType.
% We need to prove ['((forall (S:((fofType->Prop)->Prop)), ((forall (Xx:(fofType->Prop)), ((S Xx)->(cT Xx)))->(cT (fun (Xx:fofType)=> ((ex (fofType->Prop)) (fun (S0:(fofType->Prop))=> ((and (S S0)) (S0 Xx))))))))->(forall (U:(fofType->Prop)) (V:(fofType->Prop)), (((and (cT U)) (cT V))->(cT (fun (Xz:fofType)=> ((or (U Xz)) (V Xz)))))))']
% Parameter fofType:Type.
% Parameter cT:((fofType->Prop)->Prop).
% Trying to prove ((forall (S:((fofType->Prop)->Prop)), ((forall (Xx:(fofType->Prop)), ((S Xx)->(cT Xx)))->(cT (fun (Xx:fofType)=> ((ex (fofType->Prop)) (fun (S0:(fofType->Prop))=> ((and (S S0)) (S0 Xx))))))))->(forall (U:(fofType->Prop)) (V:(fofType->Prop)), (((and (cT U)) (cT V))->(cT (fun (Xz:fofType)=> ((or (U Xz)) (V Xz)))))))
% % SZS status GaveUp for /export/starexec/sandbox/benchmark/theBenchmark.p
% EOF
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