TSTP Solution File: SYN239-1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : SYN239-1 : 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 : 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 : Fri Sep  1 03:33:43 EDT 2023

% Result   : Unsatisfiable 11.79s 1.90s
% Output   : Proof 11.79s
% 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  : SYN239-1 : TPTP v8.1.2. 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
% 0.12/0.33  % Computer : n009.cluster.edu
% 0.12/0.33  % Model    : x86_64 x86_64
% 0.12/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33  % Memory   : 8042.1875MB
% 0.12/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33  % CPULimit : 300
% 0.12/0.33  % WCLimit  : 300
% 0.12/0.33  % DateTime : Sat Aug 26 18:00:35 EDT 2023
% 0.12/0.33  % CPUTime  : 
% 11.79/1.90  Command-line arguments: --ground-connectedness --complete-subsets
% 11.79/1.90  
% 11.79/1.90  % SZS status Unsatisfiable
% 11.79/1.90  
% 11.79/1.90  % SZS output start Proof
% 11.79/1.90  Take the following subset of the input axioms:
% 11.79/1.90    fof(axiom_24, axiom, l0(c)).
% 11.79/1.90    fof(prove_this, negated_conjecture, ![X]: ~m1(X, c, X)).
% 11.79/1.90    fof(rule_020, axiom, m1(c, c, c) | ~l0(c)).
% 11.79/1.90  
% 11.79/1.90  Now clausify the problem and encode Horn clauses using encoding 3 of
% 11.79/1.90  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 11.79/1.90  We repeatedly replace C & s=t => u=v by the two clauses:
% 11.79/1.90    fresh(y, y, x1...xn) = u
% 11.79/1.90    C => fresh(s, t, x1...xn) = v
% 11.79/1.90  where fresh is a fresh function symbol and x1..xn are the free
% 11.79/1.90  variables of u and v.
% 11.79/1.90  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 11.79/1.90  input problem has no model of domain size 1).
% 11.79/1.90  
% 11.79/1.90  The encoding turns the above axioms into the following unit equations and goals:
% 11.79/1.90  
% 11.79/1.90  Axiom 1 (axiom_24): l0(c) = true2.
% 11.79/1.90  Axiom 2 (rule_020): fresh416(X, X) = true2.
% 11.79/1.90  Axiom 3 (rule_020): fresh416(l0(c), true2) = m1(c, c, c).
% 11.79/1.90  
% 11.79/1.90  Goal 1 (prove_this): m1(X, c, X) = true2.
% 11.79/1.90  The goal is true when:
% 11.79/1.90    X = c
% 11.79/1.90  
% 11.79/1.90  Proof:
% 11.79/1.90    m1(c, c, c)
% 11.79/1.90  = { by axiom 3 (rule_020) R->L }
% 11.79/1.90    fresh416(l0(c), true2)
% 11.79/1.90  = { by axiom 1 (axiom_24) }
% 11.79/1.90    fresh416(true2, true2)
% 11.79/1.90  = { by axiom 2 (rule_020) }
% 11.79/1.90    true2
% 11.79/1.90  % SZS output end Proof
% 11.79/1.90  
% 11.79/1.90  RESULT: Unsatisfiable (the axioms are contradictory).
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