%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWC036-1 : 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 : n016.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 20:53:32 EDT 2023

% Result   : Unsatisfiable 3.95s 0.86s
% Output   : Proof 3.95s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.13  % Problem  : SWC036-1 : TPTP v8.1.2. Released v2.4.0.
% 0.12/0.13  % 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 : n016.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 : Mon Aug 28 16:42:29 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 3.95/0.86  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 3.95/0.86  
% 3.95/0.86  % SZS status Unsatisfiable
% 3.95/0.86  
% 3.95/0.86  % SZS output start Proof
% 3.95/0.86  Take the following subset of the input axioms:
% 3.95/0.86    fof(co1_5, negated_conjecture, nil=sk2).
% 3.95/0.86    fof(co1_6, negated_conjecture, sk2=sk4).
% 3.95/0.86    fof(co1_7, negated_conjecture, sk1=sk3).
% 3.95/0.86    fof(co1_8, negated_conjecture, nil!=sk1).
% 3.95/0.86    fof(co1_9, negated_conjecture, nil=sk3 | nil!=sk4).
% 3.95/0.86  
% 3.95/0.86  Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.95/0.86  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.95/0.86  We repeatedly replace C & s=t => u=v by the two clauses:
% 3.95/0.86    fresh(y, y, x1...xn) = u
% 3.95/0.86    C => fresh(s, t, x1...xn) = v
% 3.95/0.86  where fresh is a fresh function symbol and x1..xn are the free
% 3.95/0.86  variables of u and v.
% 3.95/0.86  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.95/0.86  input problem has no model of domain size 1).
% 3.95/0.86  
% 3.95/0.86  The encoding turns the above axioms into the following unit equations and goals:
% 3.95/0.86  
% 3.95/0.86  Axiom 1 (co1_5): nil = sk2.
% 3.95/0.86  Axiom 2 (co1_7): sk1 = sk3.
% 3.95/0.86  Axiom 3 (co1_6): sk2 = sk4.
% 3.95/0.86  Axiom 4 (co1_9): fresh14(X, X) = sk3.
% 3.95/0.86  Axiom 5 (co1_9): fresh14(nil, sk4) = nil.
% 3.95/0.86  
% 3.95/0.86  Goal 1 (co1_8): nil = sk1.
% 3.95/0.86  Proof:
% 3.95/0.86    nil
% 3.95/0.86  = { by axiom 5 (co1_9) R->L }
% 3.95/0.86    fresh14(nil, sk4)
% 3.95/0.86  = { by axiom 3 (co1_6) R->L }
% 3.95/0.86    fresh14(nil, sk2)
% 3.95/0.86  = { by axiom 1 (co1_5) R->L }
% 3.95/0.86    fresh14(nil, nil)
% 3.95/0.86  = { by axiom 4 (co1_9) }
% 3.95/0.86    sk3
% 3.95/0.86  = { by axiom 2 (co1_7) R->L }
% 3.95/0.86    sk1
% 3.95/0.86  % SZS output end Proof
% 3.95/0.86  
% 3.95/0.86  RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------