TSTP Solution File: NUN087+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUN087+2 : TPTP v8.1.2. Released v7.3.0.
% 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 : n012.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 12:51:56 EDT 2023
% Result : Theorem 0.20s 0.41s
% 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 : NUN087+2 : TPTP v8.1.2. Released v7.3.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34 % Computer : n012.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 : Sun Aug 27 09:22:55 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.20/0.41 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.20/0.41
% 0.20/0.41 % SZS status Theorem
% 0.20/0.41
% 0.20/0.41 % SZS output start Proof
% 0.20/0.41 Take the following subset of the input axioms:
% 0.20/0.41 fof(axiom_1, axiom, ?[Y24]: ![X19]: ((~r1(X19) & X19!=Y24) | (r1(X19) & X19=Y24))).
% 0.20/0.41 fof(axiom_5a, axiom, ![X5]: ?[Y8]: (?[Y17]: (r1(Y17) & r4(X5, Y17, Y8)) & ?[Y18]: (r1(Y18) & Y8=Y18))).
% 0.20/0.41 fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~r1(Y20) | Y20!=Y10) | ~r2(X7, Y10))).
% 0.20/0.41 fof(zerotimeszeroeqzero, conjecture, ?[Y1]: (?[Y2]: (r1(Y2) & r4(Y2, Y2, Y1)) & ?[Y3]: (r1(Y3) & Y1=Y3))).
% 0.20/0.41
% 0.20/0.41 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.41 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.41 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.41 fresh(y, y, x1...xn) = u
% 0.20/0.41 C => fresh(s, t, x1...xn) = v
% 0.20/0.41 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.41 variables of u and v.
% 0.20/0.41 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.41 input problem has no model of domain size 1).
% 0.20/0.41
% 0.20/0.41 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.41
% 0.20/0.41 Axiom 1 (axiom_5a): y8(X) = y18(X).
% 0.20/0.41 Axiom 2 (axiom_5a_1): r1(y17(X)) = true2.
% 0.20/0.41 Axiom 3 (axiom_5a_2): r1(y18(X)) = true2.
% 0.20/0.41 Axiom 4 (axiom_1_1): fresh10(X, X, Y) = y24.
% 0.20/0.41 Axiom 5 (axiom_1): fresh9(X, X, Y) = true2.
% 0.20/0.41 Axiom 6 (axiom_1): fresh9(X, y24, X) = r1(X).
% 0.20/0.41 Axiom 7 (axiom_1_1): fresh10(r1(X), true2, X) = X.
% 0.20/0.41 Axiom 8 (axiom_5a_3): r4(X, y17(X), y8(X)) = true2.
% 0.20/0.41
% 0.20/0.41 Lemma 9: r1(y24) = true2.
% 0.20/0.41 Proof:
% 0.20/0.41 r1(y24)
% 0.20/0.41 = { by axiom 6 (axiom_1) R->L }
% 0.20/0.41 fresh9(y24, y24, y24)
% 0.20/0.41 = { by axiom 5 (axiom_1) }
% 0.20/0.41 true2
% 0.20/0.41
% 0.20/0.41 Goal 1 (zerotimeszeroeqzero): tuple(r1(X), r1(Y), r4(X, X, Y)) = tuple(true2, true2, true2).
% 0.20/0.41 The goal is true when:
% 0.20/0.41 X = y24
% 0.20/0.41 Y = y24
% 0.20/0.41
% 0.20/0.41 Proof:
% 0.20/0.41 tuple(r1(y24), r1(y24), r4(y24, y24, y24))
% 0.20/0.41 = { by axiom 4 (axiom_1_1) R->L }
% 0.20/0.41 tuple(r1(y24), r1(y24), r4(y24, fresh10(true2, true2, y17(y24)), y24))
% 0.20/0.41 = { by axiom 2 (axiom_5a_1) R->L }
% 0.20/0.41 tuple(r1(y24), r1(y24), r4(y24, fresh10(r1(y17(y24)), true2, y17(y24)), y24))
% 0.20/0.41 = { by axiom 7 (axiom_1_1) }
% 0.20/0.41 tuple(r1(y24), r1(y24), r4(y24, y17(y24), y24))
% 0.20/0.41 = { by axiom 4 (axiom_1_1) R->L }
% 0.20/0.41 tuple(r1(y24), r1(y24), r4(y24, y17(y24), fresh10(true2, true2, y8(y24))))
% 0.20/0.41 = { by axiom 3 (axiom_5a_2) R->L }
% 0.20/0.41 tuple(r1(y24), r1(y24), r4(y24, y17(y24), fresh10(r1(y18(y24)), true2, y8(y24))))
% 0.20/0.41 = { by axiom 1 (axiom_5a) R->L }
% 0.20/0.41 tuple(r1(y24), r1(y24), r4(y24, y17(y24), fresh10(r1(y8(y24)), true2, y8(y24))))
% 0.20/0.41 = { by axiom 7 (axiom_1_1) }
% 0.20/0.41 tuple(r1(y24), r1(y24), r4(y24, y17(y24), y8(y24)))
% 0.20/0.41 = { by axiom 8 (axiom_5a_3) }
% 0.20/0.41 tuple(r1(y24), r1(y24), true2)
% 0.20/0.41 = { by lemma 9 }
% 0.20/0.41 tuple(true2, r1(y24), true2)
% 0.20/0.41 = { by lemma 9 }
% 0.20/0.41 tuple(true2, true2, true2)
% 0.20/0.41 % SZS output end Proof
% 0.20/0.41
% 0.20/0.41 RESULT: Theorem (the conjecture is true).
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