%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN243-1 : 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 : n014.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 03:33:44 EDT 2023

% Result   : Unsatisfiable 21.62s 3.21s
% Output   : Proof 21.62s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13  % Problem  : SYN243-1 : TPTP v8.1.2. Released v1.1.0.
% 0.00/0.14  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.36  % Computer : n014.cluster.edu
% 0.13/0.36  % Model    : x86_64 x86_64
% 0.13/0.36  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.36  % Memory   : 8042.1875MB
% 0.13/0.36  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.36  % CPULimit : 300
% 0.13/0.36  % WCLimit  : 300
% 0.13/0.36  % DateTime : Sat Aug 26 19:10:47 EDT 2023
% 0.13/0.36  % CPUTime  : 
% 21.62/3.21  Command-line arguments: --no-flatten-goal
% 21.62/3.21  
% 21.62/3.21  % SZS status Unsatisfiable
% 21.62/3.21  
% 21.62/3.21  % SZS output start Proof
% 21.62/3.21  Take the following subset of the input axioms:
% 21.62/3.21    fof(axiom_19, axiom, ![X, Y]: m0(X, d, Y)).
% 21.62/3.21    fof(axiom_20, axiom, l0(a)).
% 21.62/3.21    fof(prove_this, negated_conjecture, ~m1(a, d, d)).
% 21.62/3.21    fof(rule_015, axiom, ![C, D, B]: (m1(B, C, C) | (~l0(D) | ~m0(C, C, B)))).
% 21.62/3.21  
% 21.62/3.21  Now clausify the problem and encode Horn clauses using encoding 3 of
% 21.62/3.21  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 21.62/3.21  We repeatedly replace C & s=t => u=v by the two clauses:
% 21.62/3.21    fresh(y, y, x1...xn) = u
% 21.62/3.21    C => fresh(s, t, x1...xn) = v
% 21.62/3.21  where fresh is a fresh function symbol and x1..xn are the free
% 21.62/3.21  variables of u and v.
% 21.62/3.21  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 21.62/3.21  input problem has no model of domain size 1).
% 21.62/3.21  
% 21.62/3.21  The encoding turns the above axioms into the following unit equations and goals:
% 21.62/3.21  
% 21.62/3.21  Axiom 1 (axiom_20): l0(a) = true.
% 21.62/3.21  Axiom 2 (axiom_19): m0(X, d, Y) = true.
% 21.62/3.21  Axiom 3 (rule_015): fresh425(X, X, Y, Z) = m1(Y, Z, Z).
% 21.62/3.21  Axiom 4 (rule_015): fresh424(X, X, Y, Z) = true.
% 21.62/3.21  Axiom 5 (rule_015): fresh425(l0(X), true, Y, Z) = fresh424(m0(Z, Z, Y), true, Y, Z).
% 21.62/3.21  
% 21.62/3.21  Goal 1 (prove_this): m1(a, d, d) = true.
% 21.62/3.21  Proof:
% 21.62/3.21    m1(a, d, d)
% 21.62/3.21  = { by axiom 3 (rule_015) R->L }
% 21.62/3.21    fresh425(true, true, a, d)
% 21.62/3.21  = { by axiom 1 (axiom_20) R->L }
% 21.62/3.21    fresh425(l0(a), true, a, d)
% 21.62/3.21  = { by axiom 5 (rule_015) }
% 21.62/3.21    fresh424(m0(d, d, a), true, a, d)
% 21.62/3.21  = { by axiom 2 (axiom_19) }
% 21.62/3.21    fresh424(true, true, a, d)
% 21.62/3.21  = { by axiom 4 (rule_015) }
% 21.62/3.21    true
% 21.62/3.21  % SZS output end Proof
% 21.62/3.21  
% 21.62/3.21  RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------