TSTP Solution File: SET143^5 by cocATP---0.2.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : cocATP---0.2.0
% Problem  : SET143^5 : TPTP v6.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p

% Computer : n093.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 13:30:16 EDT 2014

% Result   : Theorem 0.50s
% Output   : Proof 0.50s
% Verified : 
% SZS Type : None (Parsing solution fails)
% Syntax   : Number of formulae    : 0

% Comments : 
%------------------------------------------------------------------------------
%----ERROR: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% % Problem  : SET143^5 : TPTP v6.1.0. Released v4.0.0.
% % Command  : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p
% % Computer : n093.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 09:06:46 CDT 2014
% % CPUTime  : 0.50 
% 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 0x27f6098>, <kernel.Type object at 0x2439fc8>) of role type named a_type
% Using role type
% Declaring a:Type
% FOF formula (forall (X:(a->Prop)) (Y:(a->Prop)) (Z:(a->Prop)), (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))) of role conjecture named cBOOL_PROP_67_pme
% Conjecture to prove = (forall (X:(a->Prop)) (Y:(a->Prop)) (Z:(a->Prop)), (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))):Prop
% Parameter a_DUMMY:a.
% We need to prove ['(forall (X:(a->Prop)) (Y:(a->Prop)) (Z:(a->Prop)), (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))))']
% Parameter a:Type.
% Trying to prove (forall (X:(a->Prop)) (Y:(a->Prop)) (Z:(a->Prop)), (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))))
% Found eta_expansion000:=(eta_expansion00 (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))):(((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (x:a)=> ((and ((and (X x)) (Y x))) (Z x))))
% Found (eta_expansion00 (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) as proof of (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))
% Found ((eta_expansion0 Prop) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) as proof of (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))
% Found (((eta_expansion a) Prop) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) as proof of (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))
% Found (fun (Z:(a->Prop))=> (((eta_expansion a) Prop) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))) as proof of (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))
% Found (fun (Y:(a->Prop)) (Z:(a->Prop))=> (((eta_expansion a) Prop) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))) as proof of (forall (Z:(a->Prop)), (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))))
% Found (fun (X:(a->Prop)) (Y:(a->Prop)) (Z:(a->Prop))=> (((eta_expansion a) Prop) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))) as proof of (forall (Y:(a->Prop)) (Z:(a->Prop)), (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))))
% Found (fun (X:(a->Prop)) (Y:(a->Prop)) (Z:(a->Prop))=> (((eta_expansion a) Prop) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx))))) as proof of (forall (X:(a->Prop)) (Y:(a->Prop)) (Z:(a->Prop)), (((eq (a->Prop)) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))))
% Got proof (fun (X:(a->Prop)) (Y:(a->Prop)) (Z:(a->Prop))=> (((eta_expansion a) Prop) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))))
% Time elapsed = 0.193906s
% node=14 cost=-264.000000 depth=6
% ::::::::::::::::::::::
% % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p
% % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% (fun (X:(a->Prop)) (Y:(a->Prop)) (Z:(a->Prop))=> (((eta_expansion a) Prop) (fun (Xx:a)=> ((and ((and (X Xx)) (Y Xx))) (Z Xx)))))
% % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p
% EOF
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