TSTP Solution File: NUN059+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUN059+2 : TPTP v8.1.2. Bugfixed v7.4.0.
% 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 : n031.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 12:51:47 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
% 0.07/0.13 % Problem : NUN059+2 : TPTP v8.1.2. Bugfixed v7.4.0.
% 0.07/0.14 % 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 : n031.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 : Sun Aug 27 10:02:24 EDT 2023
% 0.13/0.35 % 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(axiom_1, axiom, ?[Y24]: ![X19]: ((~r1(X19) & X19!=Y24) | (r1(X19) & X19=Y24))).
% 0.20/0.45 fof(axiom_4a, axiom, ![X4]: ?[Y9]: (?[Y16]: (r1(Y16) & r3(X4, Y16, Y9)) & Y9=X4)).
% 0.20/0.45 fof(axiom_5a, axiom, ![X5]: ?[Y8]: (?[Y17]: (r1(Y17) & r4(X5, Y17, Y8)) & ?[Y18]: (r1(Y18) & Y8=Y18))).
% 0.20/0.45 fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~r1(Y20) | Y20!=Y10) | ~r2(X7, Y10))).
% 0.20/0.45 fof(fermattothepoweroftwo, conjecture, ?[Y4, Y2, Y3, Y1]: (?[Y5]: (r4(Y2, Y2, Y5) & Y4=Y5) & ?[Y6]: (r4(Y3, Y3, Y6) & ?[Y7]: (r4(Y1, Y1, Y7) & r3(Y7, Y6, Y4))))).
% 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 (axiom_4a): y9(X) = X.
% 0.20/0.45 Axiom 2 (axiom_5a): y8(X) = y18(X).
% 0.20/0.45 Axiom 3 (axiom_4a_1): r1(y16(X)) = true2.
% 0.20/0.45 Axiom 4 (axiom_5a_1): r1(y17(X)) = true2.
% 0.20/0.45 Axiom 5 (axiom_5a_2): r1(y18(X)) = true2.
% 0.20/0.45 Axiom 6 (axiom_1_1): fresh10(X, X, Y) = y24.
% 0.20/0.45 Axiom 7 (axiom_1_1): fresh10(r1(X), true2, X) = X.
% 0.20/0.45 Axiom 8 (axiom_4a_2): r3(X, y16(X), y9(X)) = true2.
% 0.20/0.45 Axiom 9 (axiom_5a_3): r4(X, y17(X), y8(X)) = true2.
% 0.20/0.45
% 0.20/0.45 Lemma 10: r4(X, y24, y24) = true2.
% 0.20/0.45 Proof:
% 0.20/0.45 r4(X, y24, y24)
% 0.20/0.45 = { by axiom 6 (axiom_1_1) R->L }
% 0.20/0.45 r4(X, fresh10(true2, true2, y17(X)), y24)
% 0.20/0.45 = { by axiom 4 (axiom_5a_1) R->L }
% 0.20/0.45 r4(X, fresh10(r1(y17(X)), true2, y17(X)), y24)
% 0.20/0.45 = { by axiom 7 (axiom_1_1) }
% 0.20/0.45 r4(X, y17(X), y24)
% 0.20/0.45 = { by axiom 6 (axiom_1_1) R->L }
% 0.20/0.45 r4(X, y17(X), fresh10(true2, true2, y8(X)))
% 0.20/0.45 = { by axiom 5 (axiom_5a_2) R->L }
% 0.20/0.45 r4(X, y17(X), fresh10(r1(y18(X)), true2, y8(X)))
% 0.20/0.45 = { by axiom 2 (axiom_5a) R->L }
% 0.20/0.45 r4(X, y17(X), fresh10(r1(y8(X)), true2, y8(X)))
% 0.20/0.45 = { by axiom 7 (axiom_1_1) }
% 0.20/0.45 r4(X, y17(X), y8(X))
% 0.20/0.45 = { by axiom 9 (axiom_5a_3) }
% 0.20/0.45 true2
% 0.20/0.45
% 0.20/0.45 Goal 1 (fermattothepoweroftwo): tuple(r3(X, Y, Z), r4(W, W, X), r4(V, V, Y), r4(U, U, Z)) = tuple(true2, true2, true2, true2).
% 0.20/0.45 The goal is true when:
% 0.20/0.45 X = y24
% 0.20/0.45 Y = y24
% 0.20/0.45 Z = y24
% 0.20/0.45 W = y24
% 0.20/0.45 V = y24
% 0.20/0.45 U = y24
% 0.20/0.45
% 0.20/0.45 Proof:
% 0.20/0.45 tuple(r3(y24, y24, y24), r4(y24, y24, y24), r4(y24, y24, y24), r4(y24, y24, y24))
% 0.20/0.45 = { by axiom 6 (axiom_1_1) R->L }
% 0.20/0.45 tuple(r3(y24, fresh10(true2, true2, y16(y24)), y24), r4(y24, y24, y24), r4(y24, y24, y24), r4(y24, y24, y24))
% 0.20/0.45 = { by axiom 3 (axiom_4a_1) R->L }
% 0.20/0.45 tuple(r3(y24, fresh10(r1(y16(y24)), true2, y16(y24)), y24), r4(y24, y24, y24), r4(y24, y24, y24), r4(y24, y24, y24))
% 0.20/0.45 = { by axiom 7 (axiom_1_1) }
% 0.20/0.45 tuple(r3(y24, y16(y24), y24), r4(y24, y24, y24), r4(y24, y24, y24), r4(y24, y24, y24))
% 0.20/0.45 = { by axiom 1 (axiom_4a) R->L }
% 0.20/0.45 tuple(r3(y24, y16(y24), y9(y24)), r4(y24, y24, y24), r4(y24, y24, y24), r4(y24, y24, y24))
% 0.20/0.45 = { by axiom 8 (axiom_4a_2) }
% 0.20/0.45 tuple(true2, r4(y24, y24, y24), r4(y24, y24, y24), r4(y24, y24, y24))
% 0.20/0.45 = { by lemma 10 }
% 0.20/0.45 tuple(true2, r4(y24, y24, y24), true2, r4(y24, y24, y24))
% 0.20/0.45 = { by lemma 10 }
% 0.20/0.45 tuple(true2, true2, true2, r4(y24, y24, y24))
% 0.20/0.45 = { by lemma 10 }
% 0.20/0.45 tuple(true2, true2, true2, 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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