TSTP Solution File: SYN259-1 by Twee---2.4.1

View Problem - Process Solution

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% File     : Twee---2.4.1
% Problem  : SYN259-1 : TPTP v8.1.0. 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 : n005.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 Sep 29 23:22:52 EDT 2022

% Result   : Unsatisfiable 13.00s 2.02s
% Output   : Proof 13.00s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.12  % Problem  : SYN259-1 : TPTP v8.1.0. Released v1.1.0.
% 0.12/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34  % Computer : n005.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit : 300
% 0.13/0.34  % WCLimit  : 300
% 0.13/0.34  % DateTime : Mon Sep  5 02:17:03 EDT 2022
% 0.13/0.34  % CPUTime  : 
% 13.00/2.02  % SZS status Unsatisfiable
% 13.00/2.02  
% 13.00/2.02  % SZS output start Proof
% 13.00/2.02  Take the following subset of the input axioms:
% 13.00/2.02    fof(axiom_17, axiom, ![X]: q0(X, d)).
% 13.00/2.02    fof(prove_this, negated_conjecture, ![X2]: ~p1(X2, d, X2)).
% 13.00/2.02    fof(rule_067, axiom, ![E, F]: (p1(E, E, E) | ~q0(F, E))).
% 13.00/2.02  
% 13.00/2.02  Now clausify the problem and encode Horn clauses using encoding 3 of
% 13.00/2.02  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 13.00/2.02  We repeatedly replace C & s=t => u=v by the two clauses:
% 13.00/2.02    fresh(y, y, x1...xn) = u
% 13.00/2.02    C => fresh(s, t, x1...xn) = v
% 13.00/2.02  where fresh is a fresh function symbol and x1..xn are the free
% 13.00/2.02  variables of u and v.
% 13.00/2.02  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 13.00/2.02  input problem has no model of domain size 1).
% 13.00/2.02  
% 13.00/2.02  The encoding turns the above axioms into the following unit equations and goals:
% 13.00/2.02  
% 13.00/2.02  Axiom 1 (axiom_17): q0(X, d) = true2.
% 13.00/2.02  Axiom 2 (rule_067): fresh351(X, X, Y) = true2.
% 13.00/2.02  Axiom 3 (rule_067): fresh351(q0(X, Y), true2, Y) = p1(Y, Y, Y).
% 13.00/2.02  
% 13.00/2.02  Goal 1 (prove_this): p1(X, d, X) = true2.
% 13.00/2.02  The goal is true when:
% 13.00/2.02    X = d
% 13.00/2.02  
% 13.00/2.02  Proof:
% 13.00/2.02    p1(d, d, d)
% 13.00/2.02  = { by axiom 3 (rule_067) R->L }
% 13.00/2.02    fresh351(q0(X, d), true2, d)
% 13.00/2.02  = { by axiom 1 (axiom_17) }
% 13.00/2.02    fresh351(true2, true2, d)
% 13.00/2.02  = { by axiom 2 (rule_067) }
% 13.00/2.02    true2
% 13.00/2.02  % SZS output end Proof
% 13.00/2.02  
% 13.00/2.02  RESULT: Unsatisfiable (the axioms are contradictory).
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