TSTP Solution File: GRP794+1 by Crossbow---0.1
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%------------------------------------------------------------------------------
% File : Crossbow---0.1
% Problem : GRP794+1 : TPTP v8.1.0. Released v7.5.0.
% Transfm : none
% Format : tptp:raw
% Command : do_Crossbow---0.1 %s
% Computer : n032.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 600s
% DateTime : Sat Jul 16 07:45:57 EDT 2022
% Result : CounterSatisfiable 5.40s 5.63s
% Output : FiniteModel 5.40s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.11 % Problem : GRP794+1 : TPTP v8.1.0. Released v7.5.0.
% 0.04/0.12 % Command : do_Crossbow---0.1 %s
% 0.12/0.32 % Computer : n032.cluster.edu
% 0.12/0.32 % Model : x86_64 x86_64
% 0.12/0.32 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.32 % Memory : 8042.1875MB
% 0.12/0.32 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.32 % CPULimit : 300
% 0.12/0.32 % WCLimit : 600
% 0.12/0.32 % DateTime : Mon Jun 13 22:24:42 EDT 2022
% 0.12/0.32 % CPUTime :
% 0.12/0.32 /export/starexec/sandbox/solver/bin
% 0.12/0.32 crossbow.opt
% 0.12/0.32 do_Crossbow---0.1
% 0.12/0.32 eprover
% 0.12/0.32 runsolver
% 0.12/0.32 starexec_run_Crossbow---0.1
% 5.40/5.63 % SZS status CounterSatisfiable for theBenchmark.p
% 5.40/5.63 % SZS output start FiniteModel for theBenchmark.p
% 5.40/5.63 % domain size: 2
% 5.40/5.63 fof(interp, fi_domain, ![X] : (X = 0 | X = 1)).
% 5.40/5.63 fof(interp, fi_functors, b(0, 0) = 0 & b(0, 1) = 1 & b(1, 0) = 1 & b(1, 1) = 0).
% 5.40/5.63 fof(interp, fi_functors, e = 0).
% 5.40/5.63 fof(interp, fi_functors, esk1_0 = 0).
% 5.40/5.63 fof(interp, fi_functors, esk2_0 = 1).
% 5.40/5.63 fof(interp, fi_functors, esk3_0 = 0).
% 5.40/5.63 fof(interp, fi_functors, esk4_0 = 0).
% 5.40/5.63 fof(interp, fi_functors, k(0, 0) = 1 & k(0, 1) = 1 & k(1, 0) = 1 & k(1, 1) = 1).
% 5.40/5.63 fof(interp, fi_functors, m(0, 0) = 0 & m(0, 1) = 1 & m(1, 0) = 1 & m(1, 1) = 0).
% 5.40/5.63 fof(interp, fi_functors, s(0, 0) = 0 & s(0, 1) = 1 & s(1, 0) = 1 & s(1, 1) = 0).
% 5.40/5.63 % SZS output end FiniteModel for theBenchmark.p
% 5.40/5.63 % 8 lemma(s) from E
% 5.40/5.63 % cnf(cl, axiom, A = b(e, A)).
% 5.40/5.63 % cnf(cl, axiom, e = b(A, A)).
% 5.40/5.63 % cnf(cl, axiom, A = s(A, e)).
% 5.40/5.63 % cnf(cl, axiom, e = s(A, A)).
% 5.40/5.63 % cnf(cl, axiom, B = m(A, b(A, B))).
% 5.40/5.63 % cnf(cl, axiom, B = b(A, m(A, B))).
% 5.40/5.63 % cnf(cl, axiom, B = b(s(A, B), A)).
% 5.40/5.63 % cnf(cl, axiom, B = s(A, b(B, A))).
% 5.40/5.63 % 389 pred(s)
% 5.40/5.63 % 15 func(s)
% 5.40/5.63 % 2 sort(s)
% 5.40/5.63 % 430 clause(s)
% 5.40/5.63 % Instantiating 1 (5223 ms)
% 5.40/5.63 % Solving (5224 ms)
% 5.40/5.63 % Instantiating 2 (5224 ms)
% 5.40/5.63 % Solving (5280 ms)
% 5.40/5.63 %
% 5.40/5.63 % 1 model found (5284 ms)
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