TSTP Solution File: SYN168-1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : SYN168-1 : 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 : n017.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 03:33:26 EDT 2023

% Result   : Unsatisfiable 12.93s 2.12s
% Output   : Proof 12.93s
% 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  : SYN168-1 : 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
% 0.12/0.34  % Computer : n017.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit : 300
% 0.12/0.34  % WCLimit  : 300
% 0.12/0.34  % DateTime : Sat Aug 26 18:44:11 EDT 2023
% 0.12/0.34  % CPUTime  : 
% 12.93/2.12  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 12.93/2.12  
% 12.93/2.12  % SZS status Unsatisfiable
% 12.93/2.12  
% 12.93/2.13  % SZS output start Proof
% 12.93/2.13  Take the following subset of the input axioms:
% 12.93/2.13    fof(axiom_14, axiom, ![X]: p0(b, X)).
% 12.93/2.13    fof(prove_this, negated_conjecture, ![Y, X2]: ~p3(X2, Y, b)).
% 12.93/2.13    fof(rule_069, axiom, ![C, B]: (p1(B, B, C) | ~p0(C, B))).
% 12.93/2.13    fof(rule_137, axiom, ![A2, C2, B2]: (n2(A2) | ~p1(B2, C2, A2))).
% 12.93/2.13    fof(rule_244, axiom, ![H]: (p3(H, H, H) | ~n2(H))).
% 12.93/2.13  
% 12.93/2.13  Now clausify the problem and encode Horn clauses using encoding 3 of
% 12.93/2.13  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 12.93/2.13  We repeatedly replace C & s=t => u=v by the two clauses:
% 12.93/2.13    fresh(y, y, x1...xn) = u
% 12.93/2.13    C => fresh(s, t, x1...xn) = v
% 12.93/2.13  where fresh is a fresh function symbol and x1..xn are the free
% 12.93/2.13  variables of u and v.
% 12.93/2.13  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 12.93/2.13  input problem has no model of domain size 1).
% 12.93/2.13  
% 12.93/2.13  The encoding turns the above axioms into the following unit equations and goals:
% 12.93/2.13  
% 12.93/2.13  Axiom 1 (axiom_14): p0(b, X) = true2.
% 12.93/2.13  Axiom 2 (rule_137): fresh262(X, X, Y) = true2.
% 12.93/2.13  Axiom 3 (rule_244): fresh121(X, X, Y) = true2.
% 12.93/2.13  Axiom 4 (rule_069): fresh349(X, X, Y, Z) = true2.
% 12.93/2.13  Axiom 5 (rule_244): fresh121(n2(X), true2, X) = p3(X, X, X).
% 12.93/2.13  Axiom 6 (rule_069): fresh349(p0(X, Y), true2, Y, X) = p1(Y, Y, X).
% 12.93/2.13  Axiom 7 (rule_137): fresh262(p1(X, Y, Z), true2, Z) = n2(Z).
% 12.93/2.13  
% 12.93/2.13  Goal 1 (prove_this): p3(X, Y, b) = true2.
% 12.93/2.13  The goal is true when:
% 12.93/2.13    X = b
% 12.93/2.13    Y = b
% 12.93/2.13  
% 12.93/2.13  Proof:
% 12.93/2.13    p3(b, b, b)
% 12.93/2.13  = { by axiom 5 (rule_244) R->L }
% 12.93/2.13    fresh121(n2(b), true2, b)
% 12.93/2.13  = { by axiom 7 (rule_137) R->L }
% 12.93/2.13    fresh121(fresh262(p1(X, X, b), true2, b), true2, b)
% 12.93/2.13  = { by axiom 6 (rule_069) R->L }
% 12.93/2.13    fresh121(fresh262(fresh349(p0(b, X), true2, X, b), true2, b), true2, b)
% 12.93/2.13  = { by axiom 1 (axiom_14) }
% 12.93/2.13    fresh121(fresh262(fresh349(true2, true2, X, b), true2, b), true2, b)
% 12.93/2.13  = { by axiom 4 (rule_069) }
% 12.93/2.13    fresh121(fresh262(true2, true2, b), true2, b)
% 12.93/2.13  = { by axiom 2 (rule_137) }
% 12.93/2.13    fresh121(true2, true2, b)
% 12.93/2.13  = { by axiom 3 (rule_244) }
% 12.93/2.13    true2
% 12.93/2.13  % SZS output end Proof
% 12.93/2.13  
% 12.93/2.13  RESULT: Unsatisfiable (the axioms are contradictory).
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