%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN926+1 : TPTP v8.1.2. Released v3.1.0.
% 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 : n009.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 : Fri Sep  1 03:36:33 EDT 2023

% Result   : Theorem 0.15s 0.33s
% Output   : Proof 0.15s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.09/0.10  % Problem  : SYN926+1 : TPTP v8.1.2. Released v3.1.0.
% 0.09/0.10  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.10/0.30  % Computer : n009.cluster.edu
% 0.10/0.30  % Model    : x86_64 x86_64
% 0.10/0.30  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.10/0.30  % Memory   : 8042.1875MB
% 0.10/0.30  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.10/0.30  % CPULimit : 300
% 0.10/0.30  % WCLimit  : 300
% 0.10/0.30  % DateTime : Sat Aug 26 19:40:20 EDT 2023
% 0.15/0.31  % CPUTime  : 
% 0.15/0.33  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.15/0.33  
% 0.15/0.33  % SZS status Theorem
% 0.15/0.33  
% 0.15/0.33  % SZS output start Proof
% 0.15/0.33  Take the following subset of the input axioms:
% 0.15/0.33    fof(prove_this, conjecture, ?[X]: (p(X) & q(X)) => (?[X2]: p(X2) & ?[X3]: q(X3))).
% 0.15/0.33  
% 0.15/0.33  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.15/0.33  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.15/0.33  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.15/0.33    fresh(y, y, x1...xn) = u
% 0.15/0.33    C => fresh(s, t, x1...xn) = v
% 0.15/0.33  where fresh is a fresh function symbol and x1..xn are the free
% 0.15/0.33  variables of u and v.
% 0.15/0.33  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.15/0.33  input problem has no model of domain size 1).
% 0.15/0.33  
% 0.15/0.33  The encoding turns the above axioms into the following unit equations and goals:
% 0.15/0.33  
% 0.15/0.33  Axiom 1 (prove_this): p(x) = true2.
% 0.15/0.33  Axiom 2 (prove_this_1): q(x) = true2.
% 0.15/0.33  
% 0.15/0.33  Goal 1 (prove_this_2): tuple(p(X), q(Y)) = tuple(true2, true2).
% 0.15/0.33  The goal is true when:
% 0.15/0.33    X = x
% 0.15/0.33    Y = x
% 0.15/0.33  
% 0.15/0.33  Proof:
% 0.15/0.33    tuple(p(x), q(x))
% 0.15/0.33  = { by axiom 1 (prove_this) }
% 0.15/0.33    tuple(true2, q(x))
% 0.15/0.33  = { by axiom 2 (prove_this_1) }
% 0.15/0.33    tuple(true2, true2)
% 0.15/0.33  % SZS output end Proof
% 0.15/0.33  
% 0.15/0.33  RESULT: Theorem (the conjecture is true).
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