TSTP Solution File: NUN060+2 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : NUN060+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 : n009.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 0.48s
% 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
% 0.03/0.12  % Problem  : NUN060+2 : TPTP v8.1.2. Released v7.3.0.
% 0.03/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34  % Computer : n009.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.35  % CPULimit : 300
% 0.12/0.35  % WCLimit  : 300
% 0.12/0.35  % DateTime : Sun Aug 27 09:50:20 EDT 2023
% 0.12/0.35  % CPUTime  : 
% 0.20/0.48  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.48  
% 0.20/0.48  % SZS status Theorem
% 0.20/0.48  
% 0.20/0.49  % SZS output start Proof
% 0.20/0.49  Take the following subset of the input axioms:
% 0.20/0.49    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)))).
% 0.20/0.49    fof(axiom_4a, axiom, ![X4]: ?[Y9]: (?[Y16]: (r1(Y16) & r3(X4, Y16, Y9)) & Y9=X4)).
% 0.20/0.49    fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~r1(Y20) | Y20!=Y10) | ~r2(X7, Y10))).
% 0.20/0.49    fof(greq4, conjecture, ?[Y2, Y3, Y1]: (Y3=Y1 & ?[Y4_2]: (r3(Y2, Y4_2, Y3) & ?[Y5_2]: (r2(Y5_2, Y4_2) & ?[Y6]: (r2(Y6, Y5_2) & ?[Y7_2]: (r2(Y7_2, Y6) & ?[Y8]: (r1(Y8) & r2(Y8, Y7_2)))))))).
% 0.20/0.49  
% 0.20/0.49  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.49  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.49  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.49    fresh(y, y, x1...xn) = u
% 0.20/0.49    C => fresh(s, t, x1...xn) = v
% 0.20/0.49  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.49  variables of u and v.
% 0.20/0.49  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.49  input problem has no model of domain size 1).
% 0.20/0.49  
% 0.20/0.49  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.49  
% 0.20/0.49  Axiom 1 (axiom_4a_1): r1(y16(X)) = true2.
% 0.20/0.49  Axiom 2 (axiom_1a_1): r2(X, y15(Y, X)) = true2.
% 0.20/0.49  Axiom 3 (axiom_1a_3): r3(X, Y, y7(X, Y)) = true2.
% 0.20/0.49  
% 0.20/0.49  Goal 1 (greq4): tuple(r1(X), r2(Y, Z), r2(W, Y), r2(V, W), r2(X, V), r3(U, Z, T)) = tuple(true2, true2, true2, true2, true2, true2).
% 0.20/0.49  The goal is true when:
% 0.20/0.49    X = y16(X)
% 0.20/0.49    Y = y15(W, y15(Z, y15(Y, y16(X))))
% 0.20/0.49    Z = y15(V, y15(W, y15(Z, y15(Y, y16(X)))))
% 0.20/0.49    W = y15(Z, y15(Y, y16(X)))
% 0.20/0.49    V = y15(Y, y16(X))
% 0.20/0.49    U = U
% 0.20/0.49    T = y7(U, y15(V, y15(W, y15(Z, y15(Y, y16(X))))))
% 0.20/0.49  
% 0.20/0.49  Proof:
% 0.20/0.49    tuple(r1(y16(X)), r2(y15(W, y15(Z, y15(Y, y16(X)))), y15(V, y15(W, y15(Z, y15(Y, y16(X)))))), r2(y15(Z, y15(Y, y16(X))), y15(W, y15(Z, y15(Y, y16(X))))), r2(y15(Y, y16(X)), y15(Z, y15(Y, y16(X)))), r2(y16(X), y15(Y, y16(X))), r3(U, y15(V, y15(W, y15(Z, y15(Y, y16(X))))), y7(U, y15(V, y15(W, y15(Z, y15(Y, y16(X))))))))
% 0.20/0.49  = { by axiom 3 (axiom_1a_3) }
% 0.20/0.49    tuple(r1(y16(X)), r2(y15(W, y15(Z, y15(Y, y16(X)))), y15(V, y15(W, y15(Z, y15(Y, y16(X)))))), r2(y15(Z, y15(Y, y16(X))), y15(W, y15(Z, y15(Y, y16(X))))), r2(y15(Y, y16(X)), y15(Z, y15(Y, y16(X)))), r2(y16(X), y15(Y, y16(X))), true2)
% 0.20/0.49  = { by axiom 2 (axiom_1a_1) }
% 0.20/0.49    tuple(r1(y16(X)), true2, r2(y15(Z, y15(Y, y16(X))), y15(W, y15(Z, y15(Y, y16(X))))), r2(y15(Y, y16(X)), y15(Z, y15(Y, y16(X)))), r2(y16(X), y15(Y, y16(X))), true2)
% 0.20/0.49  = { by axiom 2 (axiom_1a_1) }
% 0.20/0.49    tuple(r1(y16(X)), true2, true2, r2(y15(Y, y16(X)), y15(Z, y15(Y, y16(X)))), r2(y16(X), y15(Y, y16(X))), true2)
% 0.20/0.49  = { by axiom 2 (axiom_1a_1) }
% 0.20/0.49    tuple(r1(y16(X)), true2, true2, true2, r2(y16(X), y15(Y, y16(X))), true2)
% 0.20/0.49  = { by axiom 2 (axiom_1a_1) }
% 0.20/0.49    tuple(r1(y16(X)), true2, true2, true2, true2, true2)
% 0.20/0.49  = { by axiom 1 (axiom_4a_1) }
% 0.20/0.49    tuple(true2, true2, true2, true2, true2, true2)
% 0.20/0.49  % SZS output end Proof
% 0.20/0.49  
% 0.20/0.49  RESULT: Theorem (the conjecture is true).
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