%------------------------------------------------------------------------------
% File : Twee---2.7
% Problem : SYN106-1 : TPTP v9.3.1. Released v1.1.0.
% Transfm : none
% Format : tptp:raw
% Command : run_twee /export/starexec/sandbox2/benchmark/theBenchmark.p
% Computer : n009.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8046.5625MB
% OS : Linux 6.8.0-71-generic
% CPULimit : 300s
% WCLimit : 300s
% DateTime : Tue Sep 29 01:47:05 PM UTC 2026
% Result : Unsatisfiable 5.19s 1.09s
% Output : Proof 5.19s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.05 % Problem : SYN106-1 : TPTP v9.3.1. Released v1.1.0.
% 0.00/0.07 % Command : run_twee /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.13/0.27 % Computer : n009.cluster.edu
% 0.13/0.27 % Model : x86_64 x86_64
% 0.13/0.27 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.27 % Memory : 8046.5625MB
% 0.13/0.27 % OS : Linux 6.8.0-71-generic
% 0.13/0.27 % CPULimit : 300
% 0.13/0.27 % WCLimit : 300
% 0.13/0.27 % DateTime : Mon Sep 28 15:34:14 UTC 2026
% 0.13/0.27 % CPUTime :
% 0.13/0.27 Running run_twee /export/starexec/sandbox2/benchmark/theBenchmark.p
% 5.19/1.09 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --complete-subsets --ground-joining-incomplete-limit 15 --flatten-regeneralise
% 5.19/1.09
% 5.19/1.09 % SZS status Unsatisfiable
% 5.19/1.09
% 5.19/1.09 % SZS output start Proof
% 5.19/1.09 Axiom 1 (axiom_30): n0(e, e) = true.
% 5.19/1.09 Axiom 2 (ifeq_axiom): ifeq(X, X, Y, Z) = Y.
% 5.19/1.09 Axiom 3 (rule_001): ifeq(n0(X, Y), true, k1(Y), true) = true.
% 5.19/1.09
% 5.19/1.09 Goal 1 (prove_this): k1(e) = true.
% 5.19/1.09 Proof:
% 5.19/1.09 k1(e)
% 5.19/1.09 = { by axiom 2 (ifeq_axiom) R->L }
% 5.19/1.09 ifeq(true, true, k1(e), true)
% 5.19/1.09 = { by axiom 1 (axiom_30) R->L }
% 5.19/1.09 ifeq(n0(e, e), true, k1(e), true)
% 5.19/1.09 = { by axiom 3 (rule_001) }
% 5.19/1.09 true
% 5.19/1.09 % SZS output end Proof
% 5.19/1.09
% 5.19/1.09 RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------