TSTP Solution File: LCL171-1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : LCL171-1 : 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 : Thu Aug 31 08:17:45 EDT 2023

% Result   : Unsatisfiable 0.20s 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
% 0.07/0.12  % Problem  : LCL171-1 : TPTP v8.1.2. Released v1.1.0.
% 0.07/0.13  % 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 : n009.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 : Thu Aug 24 19:16:50 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.20/0.40  Command-line arguments: --no-flatten-goal
% 0.20/0.40  
% 0.20/0.40  % SZS status Unsatisfiable
% 0.20/0.40  
% 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(axiom_1_4, axiom, ![A, B]: axiom(or(not(or(A, B)), or(B, A)))).
% 0.20/0.40    fof(prove_this, negated_conjecture, ~theorem(or(not(or(not(p), not(q))), or(not(q), not(p))))).
% 0.20/0.40    fof(rule_1, axiom, ![X]: (theorem(X) | ~axiom(X))).
% 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 (rule_1): fresh4(X, X, Y) = true.
% 0.20/0.40  Axiom 2 (rule_1): fresh4(axiom(X), true, X) = theorem(X).
% 0.20/0.40  Axiom 3 (axiom_1_4): axiom(or(not(or(X, Y)), or(Y, X))) = true.
% 0.20/0.40  
% 0.20/0.40  Goal 1 (prove_this): theorem(or(not(or(not(p), not(q))), or(not(q), not(p)))) = true.
% 0.20/0.40  Proof:
% 0.20/0.40    theorem(or(not(or(not(p), not(q))), or(not(q), not(p))))
% 0.20/0.40  = { by axiom 2 (rule_1) R->L }
% 0.20/0.40    fresh4(axiom(or(not(or(not(p), not(q))), or(not(q), not(p)))), true, or(not(or(not(p), not(q))), or(not(q), not(p))))
% 0.20/0.40  = { by axiom 3 (axiom_1_4) }
% 0.20/0.40    fresh4(true, true, or(not(or(not(p), not(q))), or(not(q), not(p))))
% 0.20/0.40  = { by axiom 1 (rule_1) }
% 0.20/0.40    true
% 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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