TSTP Solution File: ARI573_1 by SPASS+T---2.2.22
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- Process Solution
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% File : SPASS+T---2.2.22
% Problem : ARI573_1 : TPTP v8.1.0. Released v5.1.0.
% Transfm : none
% Format : tptp:raw
% Command : spasst-tptp-script %s %d
% Computer : n025.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 : Thu Jul 14 22:18:08 EDT 2022
% Result : Theorem 0.42s 1.04s
% Output : Refutation 0.42s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : ARI573_1 : TPTP v8.1.0. Released v5.1.0.
% 0.07/0.12 % Command : spasst-tptp-script %s %d
% 0.12/0.33 % Computer : n025.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 Jun 17 13:29:57 EDT 2022
% 0.12/0.33 % CPUTime :
% 0.18/0.46 % Using integer theory
% 0.42/1.04
% 0.42/1.04
% 0.42/1.04 % SZS status Theorem for /tmp/SPASST_18069_n025.cluster.edu
% 0.42/1.04
% 0.42/1.04 SPASS V 2.2.22 in combination with yices.
% 0.42/1.04 SPASS beiseite: Proof found by SMT.
% 0.42/1.04 Problem: /tmp/SPASST_18069_n025.cluster.edu
% 0.42/1.04 SPASS derived 19 clauses, backtracked 0 clauses and kept 48 clauses.
% 0.42/1.04 SPASS backtracked 1 times (1 times due to theory inconsistency).
% 0.42/1.04 SPASS allocated 6248 KBytes.
% 0.42/1.04 SPASS spent 0:00:00.01 on the problem.
% 0.42/1.04 0:00:00.00 for the input.
% 0.42/1.04 0:00:00.00 for the FLOTTER CNF translation.
% 0.42/1.04 0:00:00.00 for inferences.
% 0.42/1.04 0:00:00.00 for the backtracking.
% 0.42/1.04 0:00:00.00 for the reduction.
% 0.42/1.04 0:00:00.00 for interacting with the SMT procedure.
% 0.42/1.04
% 0.42/1.04
% 0.42/1.04 % SZS output start CNFRefutation for /tmp/SPASST_18069_n025.cluster.edu
% 0.42/1.04
% 0.42/1.04 % Here is a proof with depth 0, length 5 :
% 0.42/1.04 1[0:Inp] || -> lesseq(1,plus(times(skc5,2),uminus(skc3)))*.
% 0.42/1.04 2[0:Inp] || -> lesseq(1,plus(times(skc4,2),uminus(skc5)))*.
% 0.42/1.04 3[0:Inp] || lesseq(2,plus(plus(skc5,skc3),skc4))* -> .
% 0.42/1.04 4[0:Inp] || -> lesseq(1,plus(times(skc3,2),uminus(skc4)))*.
% 0.42/1.04 69(e)[0:ThR:4,3,2,1] || -> .
% 0.42/1.04
% 0.42/1.04 % SZS output end CNFRefutation for /tmp/SPASST_18069_n025.cluster.edu
% 0.42/1.04
% 0.42/1.04 Formulae used in the proof : fof_impl_3_ineq
% 0.42/1.04
% 0.42/1.04 SPASS+T ended
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