TSTP Solution File: SYN281-1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : SYN281-1 : 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 : n007.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 03:33:54 EDT 2023

% Result   : Unsatisfiable 13.05s 2.02s
% Output   : Proof 13.05s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem  : SYN281-1 : TPTP v8.1.2. Released v1.1.0.
% 0.13/0.13  % 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 : n007.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 17:41:56 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 13.05/2.02  Command-line arguments: --ground-connectedness --complete-subsets
% 13.05/2.02  
% 13.05/2.02  % SZS status Unsatisfiable
% 13.05/2.02  
% 13.05/2.02  % SZS output start Proof
% 13.05/2.02  Take the following subset of the input axioms:
% 13.05/2.02    fof(axiom_18, axiom, p0(c, b)).
% 13.05/2.02    fof(prove_this, negated_conjecture, ![X]: ~q1(X, c, c)).
% 13.05/2.02    fof(rule_095, axiom, ![G, F]: (q1(F, G, G) | ~p0(G, F))).
% 13.05/2.02  
% 13.05/2.02  Now clausify the problem and encode Horn clauses using encoding 3 of
% 13.05/2.02  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 13.05/2.02  We repeatedly replace C & s=t => u=v by the two clauses:
% 13.05/2.02    fresh(y, y, x1...xn) = u
% 13.05/2.02    C => fresh(s, t, x1...xn) = v
% 13.05/2.02  where fresh is a fresh function symbol and x1..xn are the free
% 13.05/2.02  variables of u and v.
% 13.05/2.02  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 13.05/2.02  input problem has no model of domain size 1).
% 13.05/2.02  
% 13.05/2.02  The encoding turns the above axioms into the following unit equations and goals:
% 13.05/2.02  
% 13.05/2.02  Axiom 1 (axiom_18): p0(c, b) = true2.
% 13.05/2.02  Axiom 2 (rule_095): fresh314(X, X, Y, Z) = true2.
% 13.05/2.02  Axiom 3 (rule_095): fresh314(p0(X, Y), true2, Y, X) = q1(Y, X, X).
% 13.05/2.02  
% 13.05/2.02  Goal 1 (prove_this): q1(X, c, c) = true2.
% 13.05/2.02  The goal is true when:
% 13.05/2.02    X = b
% 13.05/2.02  
% 13.05/2.02  Proof:
% 13.05/2.02    q1(b, c, c)
% 13.05/2.02  = { by axiom 3 (rule_095) R->L }
% 13.05/2.02    fresh314(p0(c, b), true2, b, c)
% 13.05/2.02  = { by axiom 1 (axiom_18) }
% 13.05/2.02    fresh314(true2, true2, b, c)
% 13.05/2.02  = { by axiom 2 (rule_095) }
% 13.05/2.02    true2
% 13.05/2.02  % SZS output end Proof
% 13.05/2.02  
% 13.05/2.02  RESULT: Unsatisfiable (the axioms are contradictory).
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