TSTP Solution File: SYN005-1.010 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : SYN005-1.010 : TPTP v8.1.2. Released v1.0.0.
% 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 : n028.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 : Fri Sep  1 03:32:42 EDT 2023

% Result   : Unsatisfiable 0.14s 0.40s
% 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
% 0.00/0.13  % Problem  : SYN005-1.010 : TPTP v8.1.2. Released v1.0.0.
% 0.00/0.14  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35  % Computer : n028.cluster.edu
% 0.14/0.35  % Model    : x86_64 x86_64
% 0.14/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35  % Memory   : 8042.1875MB
% 0.14/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35  % CPULimit : 300
% 0.14/0.35  % WCLimit  : 300
% 0.14/0.35  % DateTime : Sat Aug 26 21:09:38 EDT 2023
% 0.14/0.35  % CPUTime  : 
% 0.14/0.40  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.14/0.40  
% 0.14/0.40  % SZS status Unsatisfiable
% 0.14/0.40  
% 0.20/0.40  % SZS output start Proof
% 0.20/0.40  Take the following subset of the input axioms:
% 0.20/0.40    fof(disjunction, negated_conjecture, ![X1, X2, X3, X4, X5, X6, X7, X8, X9, X10]: (~p_1(X1, X2) | (~p_2(X2, X3) | (~p_3(X3, X4) | (~p_4(X4, X5) | (~p_5(X5, X6) | (~p_6(X6, X7) | (~p_7(X7, X8) | (~p_8(X8, X9) | (~p_9(X9, X10) | ~p_10(X10, X1))))))))))).
% 0.20/0.40    fof(p_1, negated_conjecture, p_1(a, a)).
% 0.20/0.40    fof(p_10, negated_conjecture, p_10(a, a)).
% 0.20/0.40    fof(p_2, negated_conjecture, p_2(a, a)).
% 0.20/0.40    fof(p_3, negated_conjecture, p_3(a, a)).
% 0.20/0.40    fof(p_4, negated_conjecture, p_4(a, a)).
% 0.20/0.40    fof(p_5, negated_conjecture, p_5(a, a)).
% 0.20/0.40    fof(p_6, negated_conjecture, p_6(a, a)).
% 0.20/0.40    fof(p_7, negated_conjecture, p_7(a, a)).
% 0.20/0.40    fof(p_8, negated_conjecture, p_8(a, a)).
% 0.20/0.40    fof(p_9, negated_conjecture, p_9(a, a)).
% 0.20/0.40  
% 0.20/0.40  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.40  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.40  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.40    fresh(y, y, x1...xn) = u
% 0.20/0.40    C => fresh(s, t, x1...xn) = v
% 0.20/0.40  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.40  variables of u and v.
% 0.20/0.40  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.40  input problem has no model of domain size 1).
% 0.20/0.40  
% 0.20/0.40  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.40  
% 0.20/0.40  Axiom 1 (p_1): p_1(a, a) = true2.
% 0.20/0.40  Axiom 2 (p_2): p_2(a, a) = true2.
% 0.20/0.40  Axiom 3 (p_3): p_3(a, a) = true2.
% 0.20/0.40  Axiom 4 (p_4): p_4(a, a) = true2.
% 0.20/0.40  Axiom 5 (p_5): p_5(a, a) = true2.
% 0.20/0.40  Axiom 6 (p_6): p_6(a, a) = true2.
% 0.20/0.40  Axiom 7 (p_7): p_7(a, a) = true2.
% 0.20/0.40  Axiom 8 (p_8): p_8(a, a) = true2.
% 0.20/0.40  Axiom 9 (p_9): p_9(a, a) = true2.
% 0.20/0.40  Axiom 10 (p_10): p_10(a, a) = true2.
% 0.20/0.40  
% 0.20/0.40  Goal 1 (disjunction): tuple(p_1(X, Y), p_2(Y, Z), p_3(Z, W), p_4(W, V), p_5(V, U), p_6(U, T), p_7(T, S), p_8(S, X2), p_9(X2, Y2), p_10(Y2, X)) = tuple(true2, true2, true2, true2, true2, true2, true2, true2, true2, true2).
% 0.20/0.40  The goal is true when:
% 0.20/0.40    X = a
% 0.20/0.40    Y = a
% 0.20/0.40    Z = a
% 0.20/0.40    W = a
% 0.20/0.40    V = a
% 0.20/0.40    U = a
% 0.20/0.40    T = a
% 0.20/0.40    S = a
% 0.20/0.40    X2 = a
% 0.20/0.40    Y2 = a
% 0.20/0.40  
% 0.20/0.40  Proof:
% 0.20/0.40    tuple(p_1(a, a), p_2(a, a), p_3(a, a), p_4(a, a), p_5(a, a), p_6(a, a), p_7(a, a), p_8(a, a), p_9(a, a), p_10(a, a))
% 0.20/0.40  = { by axiom 1 (p_1) }
% 0.20/0.40    tuple(true2, p_2(a, a), p_3(a, a), p_4(a, a), p_5(a, a), p_6(a, a), p_7(a, a), p_8(a, a), p_9(a, a), p_10(a, a))
% 0.20/0.40  = { by axiom 3 (p_3) }
% 0.20/0.40    tuple(true2, p_2(a, a), true2, p_4(a, a), p_5(a, a), p_6(a, a), p_7(a, a), p_8(a, a), p_9(a, a), p_10(a, a))
% 0.20/0.40  = { by axiom 2 (p_2) }
% 0.20/0.40    tuple(true2, true2, true2, p_4(a, a), p_5(a, a), p_6(a, a), p_7(a, a), p_8(a, a), p_9(a, a), p_10(a, a))
% 0.20/0.40  = { by axiom 5 (p_5) }
% 0.20/0.40    tuple(true2, true2, true2, p_4(a, a), true2, p_6(a, a), p_7(a, a), p_8(a, a), p_9(a, a), p_10(a, a))
% 0.20/0.40  = { by axiom 4 (p_4) }
% 0.20/0.40    tuple(true2, true2, true2, true2, true2, p_6(a, a), p_7(a, a), p_8(a, a), p_9(a, a), p_10(a, a))
% 0.20/0.40  = { by axiom 7 (p_7) }
% 0.20/0.40    tuple(true2, true2, true2, true2, true2, p_6(a, a), true2, p_8(a, a), p_9(a, a), p_10(a, a))
% 0.20/0.40  = { by axiom 6 (p_6) }
% 0.20/0.40    tuple(true2, true2, true2, true2, true2, true2, true2, p_8(a, a), p_9(a, a), p_10(a, a))
% 0.20/0.40  = { by axiom 9 (p_9) }
% 0.20/0.40    tuple(true2, true2, true2, true2, true2, true2, true2, p_8(a, a), true2, p_10(a, a))
% 0.20/0.40  = { by axiom 8 (p_8) }
% 0.20/0.40    tuple(true2, true2, true2, true2, true2, true2, true2, true2, true2, p_10(a, a))
% 0.20/0.40  = { by axiom 10 (p_10) }
% 0.20/0.40    tuple(true2, true2, true2, true2, true2, true2, true2, true2, true2, true2)
% 0.20/0.40  % SZS output end Proof
% 0.20/0.40  
% 0.20/0.40  RESULT: Unsatisfiable (the axioms are contradictory).
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