TSTP Solution File: SYN278-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN278-1 : 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 : n003.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 03:33:53 EDT 2023
% Result : Unsatisfiable 12.10s 2.03s
% Output : Proof 12.10s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : SYN278-1 : TPTP v8.1.2. Released v1.1.0.
% 0.13/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35 % Computer : n003.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Sat Aug 26 20:25:54 EDT 2023
% 0.13/0.35 % CPUTime :
% 12.10/2.03 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 12.10/2.03
% 12.10/2.03 % SZS status Unsatisfiable
% 12.10/2.03
% 12.10/2.03 % SZS output start Proof
% 12.10/2.03 Take the following subset of the input axioms:
% 12.10/2.03 fof(axiom_13, axiom, r0(e)).
% 12.10/2.03 fof(prove_this, negated_conjecture, ![X, Y]: ~q1(X, Y, e)).
% 12.10/2.03 fof(rule_116, axiom, ![E]: (q1(E, E, E) | ~r0(E))).
% 12.10/2.03
% 12.10/2.03 Now clausify the problem and encode Horn clauses using encoding 3 of
% 12.10/2.03 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 12.10/2.03 We repeatedly replace C & s=t => u=v by the two clauses:
% 12.10/2.03 fresh(y, y, x1...xn) = u
% 12.10/2.03 C => fresh(s, t, x1...xn) = v
% 12.10/2.03 where fresh is a fresh function symbol and x1..xn are the free
% 12.10/2.03 variables of u and v.
% 12.10/2.03 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 12.10/2.03 input problem has no model of domain size 1).
% 12.10/2.03
% 12.10/2.03 The encoding turns the above axioms into the following unit equations and goals:
% 12.10/2.03
% 12.10/2.03 Axiom 1 (axiom_13): r0(e) = true2.
% 12.10/2.03 Axiom 2 (rule_116): fresh287(X, X, Y) = true2.
% 12.10/2.03 Axiom 3 (rule_116): fresh287(r0(X), true2, X) = q1(X, X, X).
% 12.10/2.03
% 12.10/2.03 Goal 1 (prove_this): q1(X, Y, e) = true2.
% 12.10/2.03 The goal is true when:
% 12.10/2.03 X = e
% 12.10/2.03 Y = e
% 12.10/2.03
% 12.10/2.03 Proof:
% 12.10/2.03 q1(e, e, e)
% 12.10/2.03 = { by axiom 3 (rule_116) R->L }
% 12.10/2.03 fresh287(r0(e), true2, e)
% 12.10/2.03 = { by axiom 1 (axiom_13) }
% 12.10/2.03 fresh287(true2, true2, e)
% 12.10/2.03 = { by axiom 2 (rule_116) }
% 12.10/2.03 true2
% 12.10/2.03 % SZS output end Proof
% 12.10/2.03
% 12.10/2.03 RESULT: Unsatisfiable (the axioms are contradictory).
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