%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWC040+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 : n024.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 20:53:34 EDT 2023

% Result   : Theorem 3.65s 0.83s
% Output   : Proof 3.65s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.13  % Problem  : SWC040+1 : TPTP v8.1.2. Released v2.4.0.
% 0.03/0.14  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35  % Computer : n024.cluster.edu
% 0.14/0.35  % Model    : x86_64 x86_64
% 0.14/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35  % Memory   : 8042.1875MB
% 0.14/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35  % CPULimit : 300
% 0.14/0.35  % WCLimit  : 300
% 0.14/0.35  % DateTime : Mon Aug 28 15:30:08 EDT 2023
% 0.14/0.35  % CPUTime  : 
% 3.65/0.83  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 3.65/0.83  
% 3.65/0.83  % SZS status Theorem
% 3.65/0.83  
% 3.65/0.83  % SZS output start Proof
% 3.65/0.83  Take the following subset of the input axioms:
% 3.65/0.83    fof(co1, conjecture, ![U]: (ssList(U) => ![V]: (ssList(V) => ![W]: (ssList(W) => ![X]: (ssList(X) => (nil!=V | (V!=X | (U!=W | (nil=U | ((nil!=W & nil=X) | (![Y]: (ssItem(Y) => ![Z]: (ssList(Z) => (app(cons(Y, nil), Z)!=W | app(Z, cons(Y, nil))!=X))) & neq(X, nil)))))))))))).
% 3.65/0.83  
% 3.65/0.83  Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.65/0.83  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.65/0.83  We repeatedly replace C & s=t => u=v by the two clauses:
% 3.65/0.83    fresh(y, y, x1...xn) = u
% 3.65/0.83    C => fresh(s, t, x1...xn) = v
% 3.65/0.83  where fresh is a fresh function symbol and x1..xn are the free
% 3.65/0.83  variables of u and v.
% 3.65/0.83  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.65/0.83  input problem has no model of domain size 1).
% 3.65/0.83  
% 3.65/0.83  The encoding turns the above axioms into the following unit equations and goals:
% 3.65/0.83  
% 3.65/0.83  Axiom 1 (co1_1): u = w.
% 3.65/0.83  Axiom 2 (co1): nil = v.
% 3.65/0.83  Axiom 3 (co1_2): v = x.
% 3.65/0.83  Axiom 4 (co1_8): fresh15(X, X) = w.
% 3.65/0.83  Axiom 5 (co1_8): fresh15(nil, x) = nil.
% 3.65/0.83  
% 3.65/0.83  Goal 1 (co1_7): nil = u.
% 3.65/0.83  Proof:
% 3.65/0.83    nil
% 3.65/0.83  = { by axiom 5 (co1_8) R->L }
% 3.65/0.83    fresh15(nil, x)
% 3.65/0.83  = { by axiom 3 (co1_2) R->L }
% 3.65/0.83    fresh15(nil, v)
% 3.65/0.83  = { by axiom 2 (co1) R->L }
% 3.65/0.83    fresh15(nil, nil)
% 3.65/0.84  = { by axiom 4 (co1_8) }
% 3.65/0.84    w
% 3.65/0.84  = { by axiom 1 (co1_1) R->L }
% 3.65/0.84    u
% 3.65/0.84  % SZS output end Proof
% 3.65/0.84  
% 3.65/0.84  RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------