%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SEU123+2 : TPTP v8.1.2. Released v3.3.0.
% 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 : n001.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 17:51:05 EDT 2023

% Result   : Theorem 0.20s 0.42s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13  % Problem  : SEU123+2 : TPTP v8.1.2. Released v3.3.0.
% 0.00/0.14  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34  % Computer : n001.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 : Wed Aug 23 18:10:54 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.20/0.42  Command-line arguments: --no-flatten-goal
% 0.20/0.42  
% 0.20/0.42  % SZS status Theorem
% 0.20/0.42  
% 0.20/0.42  % SZS output start Proof
% 0.20/0.42  Take the following subset of the input axioms:
% 0.20/0.42    fof(d10_xboole_0, axiom, ![B, A2]: (A2=B <=> (subset(A2, B) & subset(B, A2)))).
% 0.20/0.42    fof(t2_xboole_1, lemma, ![A]: subset(empty_set, A)).
% 0.20/0.42    fof(t3_xboole_1, conjecture, ![A3]: (subset(A3, empty_set) => A3=empty_set)).
% 0.20/0.42  
% 0.20/0.42  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.42  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.42  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.42    fresh(y, y, x1...xn) = u
% 0.20/0.42    C => fresh(s, t, x1...xn) = v
% 0.20/0.42  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.42  variables of u and v.
% 0.20/0.42  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.42  input problem has no model of domain size 1).
% 0.20/0.42  
% 0.20/0.42  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.42  
% 0.20/0.42  Axiom 1 (t3_xboole_1): subset(a, empty_set) = true2.
% 0.20/0.42  Axiom 2 (t2_xboole_1): subset(empty_set, X) = true2.
% 0.20/0.42  Axiom 3 (d10_xboole_0_1): fresh5(X, X, Y, Z) = Y.
% 0.20/0.42  Axiom 4 (d10_xboole_0_1): fresh4(X, X, Y, Z) = Z.
% 0.20/0.42  Axiom 5 (d10_xboole_0_1): fresh5(subset(X, Y), true2, Y, X) = fresh4(subset(Y, X), true2, Y, X).
% 0.20/0.42  
% 0.20/0.42  Goal 1 (t3_xboole_1_1): a = empty_set.
% 0.20/0.42  Proof:
% 0.20/0.42    a
% 0.20/0.42  = { by axiom 3 (d10_xboole_0_1) R->L }
% 0.20/0.42    fresh5(true2, true2, a, empty_set)
% 0.20/0.42  = { by axiom 2 (t2_xboole_1) R->L }
% 0.20/0.42    fresh5(subset(empty_set, a), true2, a, empty_set)
% 0.20/0.42  = { by axiom 5 (d10_xboole_0_1) }
% 0.20/0.42    fresh4(subset(a, empty_set), true2, a, empty_set)
% 0.20/0.42  = { by axiom 1 (t3_xboole_1) }
% 0.20/0.42    fresh4(true2, true2, a, empty_set)
% 0.20/0.42  = { by axiom 4 (d10_xboole_0_1) }
% 0.20/0.42    empty_set
% 0.20/0.42  % SZS output end Proof
% 0.20/0.42  
% 0.20/0.42  RESULT: Theorem (the conjecture is true).
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