%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWC146+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 : n013.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 20:54:07 EDT 2023

% Result   : Theorem 3.16s 0.84s
% Output   : Proof 3.16s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem  : SWC146+1 : TPTP v8.1.2. Released v2.4.0.
% 0.07/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34  % Computer : n013.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit : 300
% 0.13/0.34  % WCLimit  : 300
% 0.13/0.34  % DateTime : Mon Aug 28 16:15:46 EDT 2023
% 0.19/0.34  % CPUTime  : 
% 3.16/0.84  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 3.16/0.84  
% 3.16/0.84  % SZS status Theorem
% 3.16/0.84  
% 3.16/0.84  % SZS output start Proof
% 3.16/0.84  Take the following subset of the input axioms:
% 3.16/0.84    fof(co1, conjecture, ![U]: (ssList(U) => ![V]: (ssList(V) => ![W]: (ssList(W) => ![X]: (ssList(X) => (V!=X | (U!=W | (~neq(V, nil) | (neq(X, nil) | singletonP(V)))))))))).
% 3.16/0.84  
% 3.16/0.84  Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.16/0.84  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.16/0.84  We repeatedly replace C & s=t => u=v by the two clauses:
% 3.16/0.84    fresh(y, y, x1...xn) = u
% 3.16/0.84    C => fresh(s, t, x1...xn) = v
% 3.16/0.84  where fresh is a fresh function symbol and x1..xn are the free
% 3.16/0.84  variables of u and v.
% 3.16/0.84  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.16/0.84  input problem has no model of domain size 1).
% 3.16/0.84  
% 3.16/0.84  The encoding turns the above axioms into the following unit equations and goals:
% 3.16/0.84  
% 3.16/0.84  Axiom 1 (co1_1): v = x.
% 3.16/0.84  Axiom 2 (co1_2): neq(v, nil) = true2.
% 3.16/0.84  
% 3.16/0.84  Goal 1 (co1_7): neq(x, nil) = true2.
% 3.16/0.84  Proof:
% 3.16/0.84    neq(x, nil)
% 3.16/0.84  = { by axiom 1 (co1_1) R->L }
% 3.16/0.84    neq(v, nil)
% 3.16/0.84  = { by axiom 2 (co1_2) }
% 3.16/0.84    true2
% 3.16/0.84  % SZS output end Proof
% 3.16/0.84  
% 3.16/0.84  RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------