TSTP Solution File: NUN080+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUN080+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 : n006.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 12:51:53 EDT 2023
% Result : Theorem 0.11s 0.33s
% Output : Proof 0.11s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.07 % Problem : NUN080+2 : TPTP v8.1.2. Released v7.3.0.
% 0.00/0.08 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.06/0.26 % Computer : n006.cluster.edu
% 0.06/0.26 % Model : x86_64 x86_64
% 0.06/0.26 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.06/0.26 % Memory : 8042.1875MB
% 0.06/0.26 % OS : Linux 3.10.0-693.el7.x86_64
% 0.06/0.26 % CPULimit : 300
% 0.06/0.26 % WCLimit : 300
% 0.06/0.26 % DateTime : Sun Aug 27 08:59:07 EDT 2023
% 0.06/0.26 % CPUTime :
% 0.11/0.33 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 0.11/0.33
% 0.11/0.33 % SZS status Theorem
% 0.11/0.33
% 0.11/0.33 % SZS output start Proof
% 0.11/0.33 Take the following subset of the input axioms:
% 0.11/0.33 fof(axiom_1a, axiom, ![X1, X8]: ?[Y4]: (?[Y5]: (?[Y15]: (r2(X8, Y15) & r3(X1, Y15, Y5)) & Y5=Y4) & ?[Y7]: (r2(Y7, Y4) & r3(X1, X8, Y7)))).
% 0.11/0.33 fof(axiom_4a, axiom, ![X4]: ?[Y9]: (?[Y16]: (r1(Y16) & r3(X4, Y16, Y9)) & Y9=X4)).
% 0.11/0.33 fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~r1(Y20) | Y20!=Y10) | ~r2(X7, Y10))).
% 0.11/0.33 fof(xplustwoeqy, conjecture, ?[Y2, Y3, Y1]: (Y3=Y2 & ?[Y4_2]: (r3(Y1, Y4_2, Y3) & ?[Y5_2]: (r2(Y5_2, Y4_2) & ?[Y6]: (r1(Y6) & r2(Y6, Y5_2)))))).
% 0.11/0.33
% 0.11/0.33 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.11/0.33 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.11/0.33 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.11/0.33 fresh(y, y, x1...xn) = u
% 0.11/0.33 C => fresh(s, t, x1...xn) = v
% 0.11/0.33 where fresh is a fresh function symbol and x1..xn are the free
% 0.11/0.33 variables of u and v.
% 0.11/0.33 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.11/0.33 input problem has no model of domain size 1).
% 0.11/0.33
% 0.11/0.33 The encoding turns the above axioms into the following unit equations and goals:
% 0.11/0.33
% 0.11/0.33 Axiom 1 (axiom_4a_1): r1(y16(X)) = true2.
% 0.11/0.33 Axiom 2 (axiom_1a_1): r2(X, y15(Y, X)) = true2.
% 0.11/0.33 Axiom 3 (axiom_1a_3): r3(X, Y, y7(X, Y)) = true2.
% 0.11/0.33
% 0.11/0.33 Goal 1 (xplustwoeqy): tuple(r1(X), r2(Y, Z), r2(X, Y), r3(W, Z, V)) = tuple(true2, true2, true2, true2).
% 0.11/0.33 The goal is true when:
% 0.11/0.33 X = y16(X)
% 0.11/0.33 Y = y15(Y, y16(X))
% 0.11/0.33 Z = y15(Z, y15(Y, y16(X)))
% 0.11/0.33 W = W
% 0.11/0.33 V = y7(W, y15(Z, y15(Y, y16(X))))
% 0.11/0.33
% 0.11/0.33 Proof:
% 0.11/0.33 tuple(r1(y16(X)), r2(y15(Y, y16(X)), y15(Z, y15(Y, y16(X)))), r2(y16(X), y15(Y, y16(X))), r3(W, y15(Z, y15(Y, y16(X))), y7(W, y15(Z, y15(Y, y16(X))))))
% 0.11/0.33 = { by axiom 3 (axiom_1a_3) }
% 0.11/0.33 tuple(r1(y16(X)), r2(y15(Y, y16(X)), y15(Z, y15(Y, y16(X)))), r2(y16(X), y15(Y, y16(X))), true2)
% 0.11/0.33 = { by axiom 2 (axiom_1a_1) }
% 0.11/0.33 tuple(r1(y16(X)), true2, r2(y16(X), y15(Y, y16(X))), true2)
% 0.11/0.33 = { by axiom 2 (axiom_1a_1) }
% 0.11/0.33 tuple(r1(y16(X)), true2, true2, true2)
% 0.11/0.33 = { by axiom 1 (axiom_4a_1) }
% 0.11/0.33 tuple(true2, true2, true2, true2)
% 0.11/0.33 % SZS output end Proof
% 0.11/0.33
% 0.11/0.33 RESULT: Theorem (the conjecture is true).
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