%------------------------------------------------------------------------------
% 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
%------------------------------------------------------------------------------