TSTP Solution File: SYN721-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN721-1 : TPTP v8.1.2. Released v2.5.0.
% 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 : n010.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 03:35:50 EDT 2023
% Result : Unsatisfiable 0.20s 0.38s
% Output : Proof 0.20s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SYN721-1 : TPTP v8.1.2. Released v2.5.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35 % Computer : n010.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 17:39:05 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.20/0.38 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
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% 0.20/0.38 % SZS status Unsatisfiable
% 0.20/0.38
% 0.20/0.38 % SZS output start Proof
% 0.20/0.38 Take the following subset of the input axioms:
% 0.20/0.39 fof(lx1_1, negated_conjecture, r(a, b)).
% 0.20/0.39 fof(lx1_2, negated_conjecture, ![B, A2]: (q(A2, A2) | ~r(A2, B))).
% 0.20/0.39 fof(lx1_3, negated_conjecture, ![A, C, B2]: (r(C, B2) | ~q(A, B2))).
% 0.20/0.39 fof(lx1_4, negated_conjecture, ![A3]: (~q(A3, a) | ~r(b, A3))).
% 0.20/0.39
% 0.20/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.39 fresh(y, y, x1...xn) = u
% 0.20/0.39 C => fresh(s, t, x1...xn) = v
% 0.20/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.39 variables of u and v.
% 0.20/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.39 input problem has no model of domain size 1).
% 0.20/0.39
% 0.20/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.39
% 0.20/0.39 Axiom 1 (lx1_1): r(a, b) = true2.
% 0.20/0.39 Axiom 2 (lx1_2): fresh(X, X, Y) = true2.
% 0.20/0.39 Axiom 3 (lx1_2): fresh(r(X, Y), true2, X) = q(X, X).
% 0.20/0.39 Axiom 4 (lx1_3): fresh2(X, X, Y, Z) = true2.
% 0.20/0.39 Axiom 5 (lx1_3): fresh2(q(X, Y), true2, Z, Y) = r(Z, Y).
% 0.20/0.39
% 0.20/0.39 Lemma 6: q(a, a) = true2.
% 0.20/0.39 Proof:
% 0.20/0.39 q(a, a)
% 0.20/0.39 = { by axiom 3 (lx1_2) R->L }
% 0.20/0.39 fresh(r(a, b), true2, a)
% 0.20/0.39 = { by axiom 1 (lx1_1) }
% 0.20/0.39 fresh(true2, true2, a)
% 0.20/0.39 = { by axiom 2 (lx1_2) }
% 0.20/0.39 true2
% 0.20/0.39
% 0.20/0.39 Goal 1 (lx1_4): tuple(r(b, X), q(X, a)) = tuple(true2, true2).
% 0.20/0.39 The goal is true when:
% 0.20/0.39 X = a
% 0.20/0.39
% 0.20/0.39 Proof:
% 0.20/0.39 tuple(r(b, a), q(a, a))
% 0.20/0.39 = { by axiom 5 (lx1_3) R->L }
% 0.20/0.39 tuple(fresh2(q(a, a), true2, b, a), q(a, a))
% 0.20/0.39 = { by lemma 6 }
% 0.20/0.39 tuple(fresh2(true2, true2, b, a), q(a, a))
% 0.20/0.39 = { by axiom 4 (lx1_3) }
% 0.20/0.39 tuple(true2, q(a, a))
% 0.20/0.39 = { by lemma 6 }
% 0.20/0.39 tuple(true2, true2)
% 0.20/0.39 % SZS output end Proof
% 0.20/0.39
% 0.20/0.39 RESULT: Unsatisfiable (the axioms are contradictory).
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