TSTP Solution File: SYN114-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN114-1 : TPTP v8.1.2. Released v1.1.0.
% Transfm : none
% Format : tptp:raw
% Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% Computer : n023.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 : 300s
% DateTime : Fri Sep 1 03:33:13 EDT 2023
% Result : Unsatisfiable 13.85s 2.10s
% Output : Proof 13.85s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12 % Problem : SYN114-1 : TPTP v8.1.2. Released v1.1.0.
% 0.03/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34 % Computer : n023.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Sat Aug 26 18:34:56 EDT 2023
% 0.13/0.34 % CPUTime :
% 13.85/2.10 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 13.85/2.10
% 13.85/2.10 % SZS status Unsatisfiable
% 13.85/2.10
% 13.85/2.10 % SZS output start Proof
% 13.85/2.10 Take the following subset of the input axioms:
% 13.85/2.10 fof(axiom_19, axiom, ![X, Y]: m0(X, d, Y)).
% 13.85/2.10 fof(prove_this, negated_conjecture, ![X2]: ~k3(X2, X2, b)).
% 13.85/2.10 fof(rule_107, axiom, ![A2]: (q1(e, A2, A2) | (~m0(A2, d, A2) | ~m0(e, d, A2)))).
% 13.85/2.10 fof(rule_129, axiom, ![J, A]: (k2(J, J) | ~q1(A, J, J))).
% 13.85/2.10 fof(rule_194, axiom, ![G, F]: (k3(F, F, G) | ~k2(G, F))).
% 13.85/2.10
% 13.85/2.10 Now clausify the problem and encode Horn clauses using encoding 3 of
% 13.85/2.10 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 13.85/2.10 We repeatedly replace C & s=t => u=v by the two clauses:
% 13.85/2.10 fresh(y, y, x1...xn) = u
% 13.85/2.10 C => fresh(s, t, x1...xn) = v
% 13.85/2.10 where fresh is a fresh function symbol and x1..xn are the free
% 13.85/2.10 variables of u and v.
% 13.85/2.10 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 13.85/2.10 input problem has no model of domain size 1).
% 13.85/2.10
% 13.85/2.10 The encoding turns the above axioms into the following unit equations and goals:
% 13.85/2.10
% 13.85/2.10 Axiom 1 (axiom_19): m0(X, d, Y) = true2.
% 13.85/2.10 Axiom 2 (rule_107): fresh301(X, X, Y) = q1(e, Y, Y).
% 13.85/2.10 Axiom 3 (rule_107): fresh300(X, X, Y) = true2.
% 13.85/2.10 Axiom 4 (rule_129): fresh270(X, X, Y) = true2.
% 13.85/2.10 Axiom 5 (rule_194): fresh184(X, X, Y, Z) = true2.
% 13.85/2.10 Axiom 6 (rule_107): fresh301(m0(e, d, X), true2, X) = fresh300(m0(X, d, X), true2, X).
% 13.85/2.10 Axiom 7 (rule_129): fresh270(q1(X, Y, Y), true2, Y) = k2(Y, Y).
% 13.85/2.10 Axiom 8 (rule_194): fresh184(k2(X, Y), true2, Y, X) = k3(Y, Y, X).
% 13.85/2.10
% 13.85/2.10 Goal 1 (prove_this): k3(X, X, b) = true2.
% 13.85/2.10 The goal is true when:
% 13.85/2.10 X = b
% 13.85/2.10
% 13.85/2.10 Proof:
% 13.85/2.10 k3(b, b, b)
% 13.85/2.10 = { by axiom 8 (rule_194) R->L }
% 13.85/2.10 fresh184(k2(b, b), true2, b, b)
% 13.85/2.10 = { by axiom 7 (rule_129) R->L }
% 13.85/2.10 fresh184(fresh270(q1(e, b, b), true2, b), true2, b, b)
% 13.85/2.10 = { by axiom 2 (rule_107) R->L }
% 13.85/2.10 fresh184(fresh270(fresh301(true2, true2, b), true2, b), true2, b, b)
% 13.85/2.10 = { by axiom 1 (axiom_19) R->L }
% 13.85/2.10 fresh184(fresh270(fresh301(m0(e, d, b), true2, b), true2, b), true2, b, b)
% 13.85/2.10 = { by axiom 6 (rule_107) }
% 13.85/2.10 fresh184(fresh270(fresh300(m0(b, d, b), true2, b), true2, b), true2, b, b)
% 13.85/2.10 = { by axiom 1 (axiom_19) }
% 13.85/2.10 fresh184(fresh270(fresh300(true2, true2, b), true2, b), true2, b, b)
% 13.85/2.10 = { by axiom 3 (rule_107) }
% 13.85/2.10 fresh184(fresh270(true2, true2, b), true2, b, b)
% 13.85/2.10 = { by axiom 4 (rule_129) }
% 13.85/2.10 fresh184(true2, true2, b, b)
% 13.85/2.10 = { by axiom 5 (rule_194) }
% 13.85/2.10 true2
% 13.85/2.10 % SZS output end Proof
% 13.85/2.10
% 13.85/2.10 RESULT: Unsatisfiable (the axioms are contradictory).
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