%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : KLE001+1 : TPTP v8.1.2. Released v4.0.0.
% 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 : n022.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 05:35:22 EDT 2023

% Result   : Theorem 0.21s 0.40s
% Output   : Proof 0.21s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.13  % Problem  : KLE001+1 : TPTP v8.1.2. Released v4.0.0.
% 0.13/0.14  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35  % Computer : n022.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 : Tue Aug 29 12:15:53 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.21/0.40  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.21/0.40  
% 0.21/0.40  % SZS status Theorem
% 0.21/0.40  
% 0.21/0.40  % SZS output start Proof
% 0.21/0.40  Take the following subset of the input axioms:
% 0.21/0.40    fof(additive_idempotence, axiom, ![A]: addition(A, A)=A).
% 0.21/0.40    fof(goals, conjecture, ![X0, X1, X2]: (leq(X0, X1) => leq(addition(X0, X2), addition(X0, X2)))).
% 0.21/0.40    fof(order, axiom, ![B, A2]: (leq(A2, B) <=> addition(A2, B)=B)).
% 0.21/0.40  
% 0.21/0.40  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.40  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.40  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.40    fresh(y, y, x1...xn) = u
% 0.21/0.40    C => fresh(s, t, x1...xn) = v
% 0.21/0.40  where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.40  variables of u and v.
% 0.21/0.40  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.40  input problem has no model of domain size 1).
% 0.21/0.40  
% 0.21/0.40  The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.40  
% 0.21/0.40  Axiom 1 (additive_idempotence): addition(X, X) = X.
% 0.21/0.40  Axiom 2 (order): fresh(X, X, Y, Z) = true.
% 0.21/0.40  Axiom 3 (order): fresh(addition(X, Y), Y, X, Y) = leq(X, Y).
% 0.21/0.40  
% 0.21/0.40  Goal 1 (goals_1): leq(addition(x0, x2), addition(x0, x2)) = true.
% 0.21/0.40  Proof:
% 0.21/0.40    leq(addition(x0, x2), addition(x0, x2))
% 0.21/0.40  = { by axiom 3 (order) R->L }
% 0.21/0.40    fresh(addition(addition(x0, x2), addition(x0, x2)), addition(x0, x2), addition(x0, x2), addition(x0, x2))
% 0.21/0.40  = { by axiom 1 (additive_idempotence) }
% 0.21/0.40    fresh(addition(x0, x2), addition(x0, x2), addition(x0, x2), addition(x0, x2))
% 0.21/0.40  = { by axiom 2 (order) }
% 0.21/0.40    true
% 0.21/0.40  % SZS output end Proof
% 0.21/0.40  
% 0.21/0.40  RESULT: Theorem (the conjecture is true).
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