%------------------------------------------------------------------------------
% File     : E-Darwin---1.5
% Problem  : LCL645-10.001 : TPTP v7.3.0. Released v7.3.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : e-darwin -pev TPTP -pmd true -if tptp -pl 2 -pc false -ps false %s

% Computer : n187.star.cs.uiowa.edu
% Model    : x86_64 x86_64
% CPU      : Intel(R) Xeon(R) CPU E5-2609 0 2.40GHz
% Memory   : 32218.5MB
% OS       : Linux 3.10.0-862.11.6.el7.x86_64
% CPULimit : 300s
% DateTime : Wed Feb 27 13:18:41 EST 2019

% Result   : Satisfiable 0.08s
% Output   : Model 0.08s
% Verified : 
% SZS Type : None (Parsing solution fails)
% Syntax   : Number of formulae    : 0

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.04  % Problem  : LCL645-10.001 : TPTP v7.3.0. Released v7.3.0.
% 0.00/0.05  % Command  : e-darwin -pev TPTP -pmd true -if tptp -pl 2 -pc false -ps false %s
% 0.03/0.25  % Computer : n187.star.cs.uiowa.edu
% 0.03/0.25  % Model    : x86_64 x86_64
% 0.03/0.25  % CPU      : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz
% 0.03/0.25  % Memory   : 32218.5MB
% 0.03/0.25  % OS       : Linux 3.10.0-862.11.6.el7.x86_64
% 0.03/0.25  % CPULimit : 300
% 0.03/0.25  % DateTime : Thu Feb 21 21:11:59 CST 2019
% 0.03/0.25  % CPUTime  : 
% 0.03/0.25  E-Darwin 1.5 2012/06/20 (based on Darwin 1.3)
% 0.03/0.25  
% 0.03/0.25  
% 0.03/0.25  Defaulting to tptp format.
% 0.03/0.25  Parsing /export/starexec/sandbox2/benchmark/theBenchmark.p ...
% 0.03/0.25  
% 0.03/0.25  
% 0.03/0.25  
% 0.03/0.26  Proving  ...
% 0.03/0.26  
% 0.08/0.26  % SZS status Satisfiable for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.08/0.26  
% 0.08/0.26  START OF MODEL (DIG):
% 0.08/0.26  (ifeq2(r1(sK2_main_Y, _0), true, p1(_0), true) = true)
% 0.08/0.26  (ifeq2(r1(sK1_main_Y, _0), true, p1(_0), true) = true)
% 0.08/0.26  (ifeq2(_0, _0, _1, _2) = _1)
% 0.08/0.26  (true = r1(sK3_main_X, sK2_main_Y))
% 0.08/0.26  (true = r1(sK3_main_X, sK1_main_Y))
% 0.08/0.26  (ifeq(p1(sK2_main_Y), true, a, b) = b)
% 0.08/0.26  (ifeq(p1(sK1_main_Y), true, a, b) = b)
% 0.08/0.26  (ifeq(_0, _0, _1, _2) = _1)
% 0.08/0.26  END OF MODEL
%------------------------------------------------------------------------------