TSTP Solution File: SWV386+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SWV386+1 : TPTP v8.1.2. Released v3.3.0.
% 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 : n004.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 23:04:03 EDT 2023
% Result : Theorem 0.20s 0.44s
% 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
% 0.07/0.12 % Problem : SWV386+1 : TPTP v8.1.2. Released v3.3.0.
% 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 : n004.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 : Tue Aug 29 08:06:37 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.20/0.44 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.20/0.44
% 0.20/0.44 % SZS status Theorem
% 0.20/0.44
% 0.20/0.45 % SZS output start Proof
% 0.20/0.45 Take the following subset of the input axioms:
% 0.20/0.45 fof(l22_co, conjecture, ![U, V, W]: (~check_cpq(triple(U, V, W)) => ![X]: (~check_cpq(insert_cpq(triple(U, V, W), X)) | ~ok(insert_cpq(triple(U, V, W), X))))).
% 0.20/0.45 fof(l22_l26, lemma, ![U2, V2, W2]: (~check_cpq(triple(U2, V2, W2)) => ![X2]: ~check_cpq(insert_cpq(triple(U2, V2, W2), X2)))).
% 0.20/0.45
% 0.20/0.45 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.45 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.45 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.45 fresh(y, y, x1...xn) = u
% 0.20/0.45 C => fresh(s, t, x1...xn) = v
% 0.20/0.45 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.45 variables of u and v.
% 0.20/0.45 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.45 input problem has no model of domain size 1).
% 0.20/0.45
% 0.20/0.45 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.45
% 0.20/0.45 Axiom 1 (l22_l26): fresh4(X, X, Y, Z, W) = true2.
% 0.20/0.45 Axiom 2 (l22_co): check_cpq(insert_cpq(triple(u, v, w), x)) = true2.
% 0.20/0.45 Axiom 3 (l22_l26): fresh4(check_cpq(insert_cpq(triple(X, Y, Z), W)), true2, X, Y, Z) = check_cpq(triple(X, Y, Z)).
% 0.20/0.45
% 0.20/0.45 Goal 1 (l22_co_2): check_cpq(triple(u, v, w)) = true2.
% 0.20/0.45 Proof:
% 0.20/0.45 check_cpq(triple(u, v, w))
% 0.20/0.45 = { by axiom 3 (l22_l26) R->L }
% 0.20/0.45 fresh4(check_cpq(insert_cpq(triple(u, v, w), x)), true2, u, v, w)
% 0.20/0.45 = { by axiom 2 (l22_co) }
% 0.20/0.45 fresh4(true2, true2, u, v, w)
% 0.20/0.45 = { by axiom 1 (l22_l26) }
% 0.20/0.45 true2
% 0.20/0.45 % SZS output end Proof
% 0.20/0.45
% 0.20/0.45 RESULT: Theorem (the conjecture is true).
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