%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWC203-1 : TPTP v8.1.2. Released v2.4.0.
% 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 : n007.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 20:54:24 EDT 2023

% Result   : Unsatisfiable 4.88s 0.97s
% Output   : Proof 4.88s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.10  % Problem  : SWC203-1 : TPTP v8.1.2. Released v2.4.0.
% 0.00/0.11  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.11/0.31  % Computer : n007.cluster.edu
% 0.11/0.31  % Model    : x86_64 x86_64
% 0.11/0.31  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.31  % Memory   : 8042.1875MB
% 0.11/0.31  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.11/0.31  % CPULimit : 300
% 0.11/0.31  % WCLimit  : 300
% 0.11/0.31  % DateTime : Mon Aug 28 16:34:58 EDT 2023
% 0.11/0.31  % CPUTime  : 
% 4.88/0.97  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 4.88/0.97  
% 4.88/0.97  % SZS status Unsatisfiable
% 4.88/0.97  
% 4.88/0.98  % SZS output start Proof
% 4.88/0.98  Take the following subset of the input axioms:
% 4.88/0.98    fof(co1_5, negated_conjecture, sk2=sk4).
% 4.88/0.98    fof(co1_6, negated_conjecture, sk1=sk3).
% 4.88/0.98    fof(co1_7, negated_conjecture, sk4=sk3).
% 4.88/0.98    fof(co1_8, negated_conjecture, neq(sk2, nil)).
% 4.88/0.98    fof(co1_9, negated_conjecture, ~neq(sk1, nil)).
% 4.88/0.98  
% 4.88/0.98  Now clausify the problem and encode Horn clauses using encoding 3 of
% 4.88/0.98  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 4.88/0.98  We repeatedly replace C & s=t => u=v by the two clauses:
% 4.88/0.98    fresh(y, y, x1...xn) = u
% 4.88/0.98    C => fresh(s, t, x1...xn) = v
% 4.88/0.98  where fresh is a fresh function symbol and x1..xn are the free
% 4.88/0.98  variables of u and v.
% 4.88/0.98  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 4.88/0.98  input problem has no model of domain size 1).
% 4.88/0.98  
% 4.88/0.98  The encoding turns the above axioms into the following unit equations and goals:
% 4.88/0.98  
% 4.88/0.98  Axiom 1 (co1_6): sk1 = sk3.
% 4.88/0.98  Axiom 2 (co1_5): sk2 = sk4.
% 4.88/0.98  Axiom 3 (co1_7): sk4 = sk3.
% 4.88/0.98  Axiom 4 (co1_8): neq(sk2, nil) = true2.
% 4.88/0.98  
% 4.88/0.98  Goal 1 (co1_9): neq(sk1, nil) = true2.
% 4.88/0.98  Proof:
% 4.88/0.98    neq(sk1, nil)
% 4.88/0.98  = { by axiom 1 (co1_6) }
% 4.88/0.98    neq(sk3, nil)
% 4.88/0.98  = { by axiom 3 (co1_7) R->L }
% 4.88/0.98    neq(sk4, nil)
% 4.88/0.98  = { by axiom 2 (co1_5) R->L }
% 4.88/0.98    neq(sk2, nil)
% 4.88/0.98  = { by axiom 4 (co1_8) }
% 4.88/0.98    true2
% 4.88/0.98  % SZS output end Proof
% 4.88/0.98  
% 4.88/0.98  RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------