%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : KLE059+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 : n031.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 05:35:43 EDT 2023

% Result   : Theorem 0.20s 0.40s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.13  % Problem  : KLE059+1 : TPTP v8.1.2. Released v4.0.0.
% 0.12/0.14  % 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 : n031.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 11:34:40 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.20/0.40  Command-line arguments: --no-flatten-goal
% 0.20/0.40  
% 0.20/0.40  % SZS status Theorem
% 0.20/0.40  
% 0.20/0.41  % SZS output start Proof
% 0.20/0.41  Take the following subset of the input axioms:
% 0.20/0.41    fof(domain5, axiom, ![X0, X1]: domain(addition(X0, X1))=addition(domain(X0), domain(X1))).
% 0.20/0.41    fof(goals, conjecture, ![X0_2, X1_2]: (addition(X0_2, X1_2)=X1_2 => addition(domain(X0_2), domain(X1_2))=domain(X1_2))).
% 0.20/0.41  
% 0.20/0.41  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.41  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.41  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.41    fresh(y, y, x1...xn) = u
% 0.20/0.41    C => fresh(s, t, x1...xn) = v
% 0.20/0.41  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.41  variables of u and v.
% 0.20/0.41  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.41  input problem has no model of domain size 1).
% 0.20/0.41  
% 0.20/0.41  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.41  
% 0.20/0.41  Axiom 1 (goals): addition(x0, x1) = x1.
% 0.20/0.41  Axiom 2 (domain5): domain(addition(X, Y)) = addition(domain(X), domain(Y)).
% 0.20/0.41  
% 0.20/0.41  Goal 1 (goals_1): addition(domain(x0), domain(x1)) = domain(x1).
% 0.20/0.41  Proof:
% 0.20/0.41    addition(domain(x0), domain(x1))
% 0.20/0.41  = { by axiom 2 (domain5) R->L }
% 0.20/0.41    domain(addition(x0, x1))
% 0.20/0.41  = { by axiom 1 (goals) }
% 0.20/0.41    domain(x1)
% 0.20/0.41  % SZS output end Proof
% 0.20/0.41  
% 0.20/0.41  RESULT: Theorem (the conjecture is true).
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