TSTP Solution File: SWC102+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SWC102+1 : TPTP v8.1.2. Released v2.4.0.
% 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 : n009.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 : Thu Aug 31 20:53:52 EDT 2023
% Result : Theorem 3.17s 0.80s
% Output : Proof 3.17s
% 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 : SWC102+1 : TPTP v8.1.2. Released v2.4.0.
% 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 : n009.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 : Mon Aug 28 16:38:50 EDT 2023
% 0.13/0.34 % CPUTime :
% 3.17/0.80 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 3.17/0.80
% 3.17/0.80 % SZS status Theorem
% 3.17/0.80
% 3.17/0.80 % SZS output start Proof
% 3.17/0.80 Take the following subset of the input axioms:
% 3.17/0.80 fof(ax42, axiom, ![U]: (ssList(U) => frontsegP(U, U))).
% 3.17/0.80 fof(co1, conjecture, ![U2]: (ssList(U2) => ![V]: (ssList(V) => ![W]: (ssList(W) => ![X]: (ssList(X) => (V!=X | (U2!=W | (X!=W | (~neq(V, nil) | (neq(U2, nil) & frontsegP(V, U2))))))))))).
% 3.17/0.80
% 3.17/0.80 Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.17/0.80 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.17/0.80 We repeatedly replace C & s=t => u=v by the two clauses:
% 3.17/0.80 fresh(y, y, x1...xn) = u
% 3.17/0.80 C => fresh(s, t, x1...xn) = v
% 3.17/0.80 where fresh is a fresh function symbol and x1..xn are the free
% 3.17/0.80 variables of u and v.
% 3.17/0.80 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.17/0.80 input problem has no model of domain size 1).
% 3.17/0.80
% 3.17/0.80 The encoding turns the above axioms into the following unit equations and goals:
% 3.17/0.80
% 3.17/0.80 Axiom 1 (co1): u = w.
% 3.17/0.80 Axiom 2 (co1_1): v = x.
% 3.17/0.80 Axiom 3 (co1_2): x = w.
% 3.17/0.80 Axiom 4 (co1_4): ssList(u) = true2.
% 3.17/0.80 Axiom 5 (co1_3): neq(v, nil) = true2.
% 3.17/0.80 Axiom 6 (ax42): fresh58(X, X, Y) = true2.
% 3.17/0.80 Axiom 7 (ax42): fresh58(ssList(X), true2, X) = frontsegP(X, X).
% 3.17/0.80
% 3.17/0.80 Lemma 8: v = u.
% 3.17/0.80 Proof:
% 3.17/0.80 v
% 3.17/0.80 = { by axiom 2 (co1_1) }
% 3.17/0.80 x
% 3.17/0.80 = { by axiom 3 (co1_2) }
% 3.17/0.80 w
% 3.17/0.80 = { by axiom 1 (co1) R->L }
% 3.17/0.80 u
% 3.17/0.80
% 3.17/0.80 Goal 1 (co1_8): tuple2(neq(u, nil), frontsegP(v, u)) = tuple2(true2, true2).
% 3.17/0.80 Proof:
% 3.17/0.80 tuple2(neq(u, nil), frontsegP(v, u))
% 3.17/0.80 = { by lemma 8 R->L }
% 3.17/0.80 tuple2(neq(v, nil), frontsegP(v, u))
% 3.17/0.80 = { by axiom 5 (co1_3) }
% 3.17/0.80 tuple2(true2, frontsegP(v, u))
% 3.17/0.80 = { by lemma 8 }
% 3.17/0.80 tuple2(true2, frontsegP(u, u))
% 3.17/0.80 = { by axiom 7 (ax42) R->L }
% 3.17/0.80 tuple2(true2, fresh58(ssList(u), true2, u))
% 3.17/0.80 = { by axiom 4 (co1_4) }
% 3.17/0.80 tuple2(true2, fresh58(true2, true2, u))
% 3.17/0.80 = { by axiom 6 (ax42) }
% 3.17/0.80 tuple2(true2, true2)
% 3.17/0.80 % SZS output end Proof
% 3.17/0.80
% 3.17/0.80 RESULT: Theorem (the conjecture is true).
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