TSTP Solution File: SYN153-1 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN153-1 : 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 : n011.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 03:33:23 EDT 2023

% Result   : Unsatisfiable 13.39s 2.13s
% Output   : Proof 13.39s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.13  % Problem  : SYN153-1 : TPTP v8.1.2. Released v1.1.0.
% 0.13/0.14  % 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 : n011.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 18:04:39 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 13.39/2.13  Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 13.39/2.13  
% 13.39/2.13  % SZS status Unsatisfiable
% 13.39/2.13  
% 13.39/2.13  % SZS output start Proof
% 13.39/2.13  Take the following subset of the input axioms:
% 13.39/2.13    fof(axiom_14, axiom, ![X]: p0(b, X)).
% 13.39/2.13    fof(axiom_5, axiom, s0(b)).
% 13.39/2.13    fof(prove_this, negated_conjecture, ~n3(a)).
% 13.39/2.13    fof(rule_029, axiom, ![I, H]: (m1(H, I, H) | (~p0(H, I) | ~s0(H)))).
% 13.39/2.13    fof(rule_176, axiom, ![D, E]: (p2(D, E, D) | ~m1(E, D, E))).
% 13.39/2.13    fof(rule_240, axiom, ![F, D2, E2]: (n3(D2) | ~p2(E2, F, D2))).
% 13.39/2.13  
% 13.39/2.13  Now clausify the problem and encode Horn clauses using encoding 3 of
% 13.39/2.13  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 13.39/2.13  We repeatedly replace C & s=t => u=v by the two clauses:
% 13.39/2.13    fresh(y, y, x1...xn) = u
% 13.39/2.13    C => fresh(s, t, x1...xn) = v
% 13.39/2.13  where fresh is a fresh function symbol and x1..xn are the free
% 13.39/2.13  variables of u and v.
% 13.39/2.13  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 13.39/2.13  input problem has no model of domain size 1).
% 13.39/2.13  
% 13.39/2.13  The encoding turns the above axioms into the following unit equations and goals:
% 13.39/2.13  
% 13.39/2.13  Axiom 1 (axiom_5): s0(b) = true.
% 13.39/2.13  Axiom 2 (axiom_14): p0(b, X) = true.
% 13.39/2.13  Axiom 3 (rule_240): fresh127(X, X, Y) = true.
% 13.39/2.13  Axiom 4 (rule_029): fresh404(X, X, Y, Z) = m1(Y, Z, Y).
% 13.39/2.13  Axiom 5 (rule_029): fresh403(X, X, Y, Z) = true.
% 13.39/2.13  Axiom 6 (rule_176): fresh208(X, X, Y, Z) = true.
% 13.39/2.13  Axiom 7 (rule_029): fresh404(p0(X, Y), true, X, Y) = fresh403(s0(X), true, X, Y).
% 13.39/2.13  Axiom 8 (rule_240): fresh127(p2(X, Y, Z), true, Z) = n3(Z).
% 13.39/2.13  Axiom 9 (rule_176): fresh208(m1(X, Y, X), true, Y, X) = p2(Y, X, Y).
% 13.39/2.13  
% 13.39/2.13  Goal 1 (prove_this): n3(a) = true.
% 13.39/2.13  Proof:
% 13.39/2.13    n3(a)
% 13.39/2.13  = { by axiom 8 (rule_240) R->L }
% 13.39/2.13    fresh127(p2(a, b, a), true, a)
% 13.39/2.13  = { by axiom 9 (rule_176) R->L }
% 13.39/2.13    fresh127(fresh208(m1(b, a, b), true, a, b), true, a)
% 13.39/2.13  = { by axiom 4 (rule_029) R->L }
% 13.39/2.13    fresh127(fresh208(fresh404(true, true, b, a), true, a, b), true, a)
% 13.39/2.13  = { by axiom 2 (axiom_14) R->L }
% 13.39/2.13    fresh127(fresh208(fresh404(p0(b, a), true, b, a), true, a, b), true, a)
% 13.39/2.13  = { by axiom 7 (rule_029) }
% 13.39/2.13    fresh127(fresh208(fresh403(s0(b), true, b, a), true, a, b), true, a)
% 13.39/2.13  = { by axiom 1 (axiom_5) }
% 13.39/2.13    fresh127(fresh208(fresh403(true, true, b, a), true, a, b), true, a)
% 13.39/2.13  = { by axiom 5 (rule_029) }
% 13.39/2.13    fresh127(fresh208(true, true, a, b), true, a)
% 13.39/2.13  = { by axiom 6 (rule_176) }
% 13.39/2.13    fresh127(true, true, a)
% 13.39/2.13  = { by axiom 3 (rule_240) }
% 13.39/2.13    true
% 13.39/2.13  % SZS output end Proof
% 13.39/2.13  
% 13.39/2.13  RESULT: Unsatisfiable (the axioms are contradictory).
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