%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN256-1 : 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 : n006.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 03:33:48 EDT 2023

% Result   : Unsatisfiable 12.35s 1.99s
% Output   : Proof 12.35s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.14  % Problem  : SYN256-1 : TPTP v8.1.2. Released v1.1.0.
% 0.15/0.15  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.17/0.37  % Computer : n006.cluster.edu
% 0.17/0.37  % Model    : x86_64 x86_64
% 0.17/0.37  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.17/0.37  % Memory   : 8042.1875MB
% 0.17/0.37  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.17/0.37  % CPULimit : 300
% 0.17/0.37  % WCLimit  : 300
% 0.17/0.37  % DateTime : Sat Aug 26 19:22:22 EDT 2023
% 0.23/0.37  % CPUTime  : 
% 12.35/1.99  Command-line arguments: --ground-connectedness --complete-subsets
% 12.35/1.99  
% 12.35/1.99  % SZS status Unsatisfiable
% 12.35/1.99  
% 12.35/1.99  % SZS output start Proof
% 12.35/1.99  Take the following subset of the input axioms:
% 12.35/1.99    fof(axiom_18, axiom, p0(c, b)).
% 12.35/1.99    fof(prove_this, negated_conjecture, ~n2(c)).
% 12.35/1.99    fof(rule_069, axiom, ![C, B]: (p1(B, B, C) | ~p0(C, B))).
% 12.35/1.99    fof(rule_137, axiom, ![A2, C2, B2]: (n2(A2) | ~p1(B2, C2, A2))).
% 12.35/1.99  
% 12.35/1.99  Now clausify the problem and encode Horn clauses using encoding 3 of
% 12.35/1.99  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 12.35/1.99  We repeatedly replace C & s=t => u=v by the two clauses:
% 12.35/1.99    fresh(y, y, x1...xn) = u
% 12.35/1.99    C => fresh(s, t, x1...xn) = v
% 12.35/1.99  where fresh is a fresh function symbol and x1..xn are the free
% 12.35/1.99  variables of u and v.
% 12.35/1.99  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 12.35/1.99  input problem has no model of domain size 1).
% 12.35/1.99  
% 12.35/1.99  The encoding turns the above axioms into the following unit equations and goals:
% 12.35/1.99  
% 12.35/1.99  Axiom 1 (axiom_18): p0(c, b) = true.
% 12.35/1.99  Axiom 2 (rule_137): fresh262(X, X, Y) = true.
% 12.35/1.99  Axiom 3 (rule_069): fresh349(X, X, Y, Z) = true.
% 12.35/1.99  Axiom 4 (rule_069): fresh349(p0(X, Y), true, Y, X) = p1(Y, Y, X).
% 12.35/1.99  Axiom 5 (rule_137): fresh262(p1(X, Y, Z), true, Z) = n2(Z).
% 12.35/1.99  
% 12.35/1.99  Goal 1 (prove_this): n2(c) = true.
% 12.35/1.99  Proof:
% 12.35/1.99    n2(c)
% 12.35/1.99  = { by axiom 5 (rule_137) R->L }
% 12.35/1.99    fresh262(p1(b, b, c), true, c)
% 12.35/1.99  = { by axiom 4 (rule_069) R->L }
% 12.35/1.99    fresh262(fresh349(p0(c, b), true, b, c), true, c)
% 12.35/1.99  = { by axiom 1 (axiom_18) }
% 12.35/1.99    fresh262(fresh349(true, true, b, c), true, c)
% 12.35/1.99  = { by axiom 3 (rule_069) }
% 12.35/1.99    fresh262(true, true, c)
% 12.35/1.99  = { by axiom 2 (rule_137) }
% 12.35/1.99    true
% 12.35/1.99  % SZS output end Proof
% 12.35/1.99  
% 12.35/1.99  RESULT: Unsatisfiable (the axioms are contradictory).
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