TSTP Solution File: SYN279-1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : SYN279-1 : TPTP v8.1.2. Released v1.1.0.
% 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 : n004.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 03:33:53 EDT 2023

% Result   : Unsatisfiable 31.96s 4.51s
% Output   : Proof 31.96s
% 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  : SYN279-1 : TPTP v8.1.2. Released v1.1.0.
% 0.00/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35  % Computer : n004.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 : Sat Aug 26 21:48:23 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 31.96/4.51  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 31.96/4.51  
% 31.96/4.51  % SZS status Unsatisfiable
% 31.96/4.51  
% 31.96/4.51  % SZS output start Proof
% 31.96/4.51  Take the following subset of the input axioms:
% 31.96/4.51    fof(axiom_14, axiom, ![X]: p0(b, X)).
% 31.96/4.51    fof(axiom_37, axiom, n0(b, a)).
% 31.96/4.51    fof(prove_this, negated_conjecture, ![X2]: ~q1(X2, a, c)).
% 31.96/4.51    fof(rule_092, axiom, ![J, C, B, A2]: (q1(J, A2, J) | (~n0(B, A2) | ~p0(C, J)))).
% 31.96/4.51  
% 31.96/4.51  Now clausify the problem and encode Horn clauses using encoding 3 of
% 31.96/4.51  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 31.96/4.51  We repeatedly replace C & s=t => u=v by the two clauses:
% 31.96/4.51    fresh(y, y, x1...xn) = u
% 31.96/4.51    C => fresh(s, t, x1...xn) = v
% 31.96/4.51  where fresh is a fresh function symbol and x1..xn are the free
% 31.96/4.51  variables of u and v.
% 31.96/4.51  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 31.96/4.51  input problem has no model of domain size 1).
% 31.96/4.51  
% 31.96/4.51  The encoding turns the above axioms into the following unit equations and goals:
% 31.96/4.51  
% 31.96/4.51  Axiom 1 (axiom_14): p0(b, X) = true2.
% 31.96/4.51  Axiom 2 (axiom_37): n0(b, a) = true2.
% 31.96/4.51  Axiom 3 (rule_092): fresh317(X, X, Y, Z) = true2.
% 31.96/4.51  Axiom 4 (rule_092): fresh318(X, X, Y, Z, W) = q1(Y, Z, Y).
% 31.96/4.51  Axiom 5 (rule_092): fresh318(p0(X, Y), true2, Y, Z, W) = fresh317(n0(W, Z), true2, Y, Z).
% 31.96/4.51  
% 31.96/4.51  Goal 1 (prove_this): q1(X, a, c) = true2.
% 31.96/4.51  The goal is true when:
% 31.96/4.51    X = c
% 31.96/4.51  
% 31.96/4.51  Proof:
% 31.96/4.51    q1(c, a, c)
% 31.96/4.51  = { by axiom 4 (rule_092) R->L }
% 31.96/4.51    fresh318(true2, true2, c, a, b)
% 31.96/4.51  = { by axiom 1 (axiom_14) R->L }
% 31.96/4.51    fresh318(p0(b, c), true2, c, a, b)
% 31.96/4.51  = { by axiom 5 (rule_092) }
% 31.96/4.51    fresh317(n0(b, a), true2, c, a)
% 31.96/4.51  = { by axiom 2 (axiom_37) }
% 31.96/4.51    fresh317(true2, true2, c, a)
% 31.96/4.51  = { by axiom 3 (rule_092) }
% 31.96/4.51    true2
% 31.96/4.51  % SZS output end Proof
% 31.96/4.51  
% 31.96/4.51  RESULT: Unsatisfiable (the axioms are contradictory).
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