TSTP Solution File: DAT029_1 by SPASS+T---2.2.22
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% File : SPASS+T---2.2.22
% Problem : DAT029_1 : TPTP v8.1.0. Released v5.0.0.
% Transfm : none
% Format : tptp:raw
% Command : spasst-tptp-script %s %d
% Computer : n004.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 01:31:51 EDT 2022
% Result : Theorem 0.75s 1.02s
% Output : Refutation 0.75s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12 % Problem : DAT029_1 : TPTP v8.1.0. Released v5.0.0.
% 0.06/0.12 % Command : spasst-tptp-script %s %d
% 0.12/0.33 % Computer : n004.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 600
% 0.12/0.33 % DateTime : Fri Jul 1 19:33:23 EDT 2022
% 0.12/0.33 % CPUTime :
% 0.19/0.46 % Using integer theory
% 0.75/1.02
% 0.75/1.02
% 0.75/1.02 % SZS status Theorem for /tmp/SPASST_32053_n004.cluster.edu
% 0.75/1.02
% 0.75/1.02 SPASS V 2.2.22 in combination with yices.
% 0.75/1.02 SPASS beiseite: Proof found by SMT.
% 0.75/1.02 Problem: /tmp/SPASST_32053_n004.cluster.edu
% 0.75/1.02 SPASS derived 90 clauses, backtracked 0 clauses and kept 87 clauses.
% 0.75/1.02 SPASS backtracked 1 times (1 times due to theory inconsistency).
% 0.75/1.02 SPASS allocated 6327 KBytes.
% 0.75/1.02 SPASS spent 0:00:00.02 on the problem.
% 0.75/1.02 0:00:00.00 for the input.
% 0.75/1.02 0:00:00.01 for the FLOTTER CNF translation.
% 0.75/1.02 0:00:00.00 for inferences.
% 0.75/1.02 0:00:00.00 for the backtracking.
% 0.75/1.02 0:00:00.00 for the reduction.
% 0.75/1.02 0:00:00.00 for interacting with the SMT procedure.
% 0.75/1.02
% 0.75/1.02
% 0.75/1.02 % SZS output start CNFRefutation for /tmp/SPASST_32053_n004.cluster.edu
% 0.75/1.02
% 0.75/1.02 % Here is a proof with depth 0, length 6 :
% 0.75/1.02 5[0:Inp] || -> in(skc7,skc5)*.
% 0.75/1.02 7[0:Inp] || greater(skc7,2)* -> .
% 0.75/1.02 10[0:Inp] || in(U,skc4)* -> greater(U,0).
% 0.75/1.02 12[0:Inp] || in(U,skc5)* -> in(skf1(V),skc4)*.
% 0.75/1.02 18[0:Inp] || in(U,skc5)* -> greater(minus(times(2,U),times(5,skf1(U))),0)*.
% 0.75/1.02 165(e)[0:ThR:18,12,10,7,5] || -> .
% 0.75/1.02
% 0.75/1.02 % SZS output end CNFRefutation for /tmp/SPASST_32053_n004.cluster.edu
% 0.75/1.02
% 0.75/1.02 Formulae used in the proof : fof_co1 fof_ax1 fof_add_type fof_ax2 fof_ax5
% 0.75/1.02
% 0.75/1.02 SPASS+T ended
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