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