TSTP Solution File: SWC121+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC121+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 : n018.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 20:53:58 EDT 2023
% Result : Theorem 3.57s 0.83s
% Output : Proof 3.57s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SWC121+1 : TPTP v8.1.2. Released v2.4.0.
% 0.07/0.13 % 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 : n018.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 16:19:47 EDT 2023
% 0.13/0.34 % CPUTime :
% 3.57/0.83 Command-line arguments: --no-flatten-goal
% 3.57/0.83
% 3.57/0.83 % SZS status Theorem
% 3.57/0.83
% 3.57/0.83 % SZS output start Proof
% 3.57/0.83 Take the following subset of the input axioms:
% 3.57/0.83 fof(co1, conjecture, ![U]: (ssList(U) => ![V]: (ssList(V) => ![W]: (ssList(W) => ![X]: (ssList(X) => (V!=X | (U!=W | ((~neq(V, nil) | (~neq(W, nil) | (~segmentP(X, W) | (neq(U, nil) & segmentP(V, U))))) & (~neq(V, nil) | neq(X, nil)))))))))).
% 3.57/0.83
% 3.57/0.83 Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.57/0.83 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.57/0.83 We repeatedly replace C & s=t => u=v by the two clauses:
% 3.57/0.83 fresh(y, y, x1...xn) = u
% 3.57/0.83 C => fresh(s, t, x1...xn) = v
% 3.57/0.83 where fresh is a fresh function symbol and x1..xn are the free
% 3.57/0.83 variables of u and v.
% 3.57/0.83 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.57/0.83 input problem has no model of domain size 1).
% 3.57/0.83
% 3.57/0.83 The encoding turns the above axioms into the following unit equations and goals:
% 3.57/0.83
% 3.57/0.83 Axiom 1 (co1): u = w.
% 3.57/0.83 Axiom 2 (co1_1): v = x.
% 3.57/0.83 Axiom 3 (co1_12): fresh15(X, X) = true2.
% 3.57/0.83 Axiom 4 (co1_13): fresh14(X, X) = true2.
% 3.57/0.83 Axiom 5 (co1_2): neq(v, nil) = true2.
% 3.57/0.83 Axiom 6 (co1_12): fresh15(neq(x, nil), true2) = neq(w, nil).
% 3.57/0.83 Axiom 7 (co1_13): fresh14(neq(x, nil), true2) = segmentP(x, w).
% 3.57/0.83
% 3.57/0.83 Goal 1 (co1_9): tuple2(neq(u, nil), neq(x, nil), segmentP(v, u)) = tuple2(true2, true2, true2).
% 3.57/0.83 Proof:
% 3.57/0.83 tuple2(neq(u, nil), neq(x, nil), segmentP(v, u))
% 3.57/0.83 = { by axiom 1 (co1) }
% 3.57/0.83 tuple2(neq(w, nil), neq(x, nil), segmentP(v, u))
% 3.57/0.83 = { by axiom 1 (co1) }
% 3.57/0.83 tuple2(neq(w, nil), neq(x, nil), segmentP(v, w))
% 3.57/0.83 = { by axiom 2 (co1_1) R->L }
% 3.57/0.83 tuple2(neq(w, nil), neq(v, nil), segmentP(v, w))
% 3.57/0.83 = { by axiom 5 (co1_2) }
% 3.57/0.83 tuple2(neq(w, nil), true2, segmentP(v, w))
% 3.57/0.83 = { by axiom 6 (co1_12) R->L }
% 3.57/0.83 tuple2(fresh15(neq(x, nil), true2), true2, segmentP(v, w))
% 3.57/0.83 = { by axiom 2 (co1_1) R->L }
% 3.57/0.83 tuple2(fresh15(neq(v, nil), true2), true2, segmentP(v, w))
% 3.57/0.83 = { by axiom 5 (co1_2) }
% 3.57/0.83 tuple2(fresh15(true2, true2), true2, segmentP(v, w))
% 3.57/0.83 = { by axiom 3 (co1_12) }
% 3.57/0.83 tuple2(true2, true2, segmentP(v, w))
% 3.57/0.83 = { by axiom 2 (co1_1) }
% 3.57/0.83 tuple2(true2, true2, segmentP(x, w))
% 3.57/0.83 = { by axiom 7 (co1_13) R->L }
% 3.57/0.83 tuple2(true2, true2, fresh14(neq(x, nil), true2))
% 3.57/0.83 = { by axiom 2 (co1_1) R->L }
% 3.57/0.83 tuple2(true2, true2, fresh14(neq(v, nil), true2))
% 3.57/0.83 = { by axiom 5 (co1_2) }
% 3.57/0.83 tuple2(true2, true2, fresh14(true2, true2))
% 3.57/0.83 = { by axiom 4 (co1_13) }
% 3.57/0.83 tuple2(true2, true2, true2)
% 3.57/0.83 % SZS output end Proof
% 3.57/0.83
% 3.57/0.83 RESULT: Theorem (the conjecture is true).
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