%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN969+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 : n011.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 03:36:41 EDT 2023

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

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.14  % Problem  : SYN969+1 : TPTP v8.1.2. Released v3.1.0.
% 0.12/0.15  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.36  % Computer : n011.cluster.edu
% 0.14/0.36  % Model    : x86_64 x86_64
% 0.14/0.36  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.36  % Memory   : 8042.1875MB
% 0.14/0.36  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.36  % CPULimit : 300
% 0.14/0.36  % WCLimit  : 300
% 0.14/0.36  % DateTime : Sat Aug 26 17:08:09 EDT 2023
% 0.14/0.36  % CPUTime  : 
% 0.21/0.40  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 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(prove_this, conjecture, ![B]: ((![X]: (p(X) => q(X)) & r(B)) => (![Y]: (r(Y) => p(Y)) => q(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 (prove_this): r(b) = true.
% 0.21/0.40  Axiom 2 (prove_this_1): fresh(X, X, Y) = true.
% 0.21/0.40  Axiom 3 (prove_this_3): fresh2(X, X, Y) = true.
% 0.21/0.40  Axiom 4 (prove_this_1): fresh(p(X), true, X) = q(X).
% 0.21/0.40  Axiom 5 (prove_this_3): fresh2(r(X), true, X) = p(X).
% 0.21/0.40  
% 0.21/0.40  Goal 1 (prove_this_2): q(b) = true.
% 0.21/0.40  Proof:
% 0.21/0.40    q(b)
% 0.21/0.40  = { by axiom 4 (prove_this_1) R->L }
% 0.21/0.40    fresh(p(b), true, b)
% 0.21/0.40  = { by axiom 5 (prove_this_3) R->L }
% 0.21/0.40    fresh(fresh2(r(b), true, b), true, b)
% 0.21/0.40  = { by axiom 1 (prove_this) }
% 0.21/0.40    fresh(fresh2(true, true, b), true, b)
% 0.21/0.40  = { by axiom 3 (prove_this_3) }
% 0.21/0.40    fresh(true, true, b)
% 0.21/0.40  = { by axiom 2 (prove_this_1) }
% 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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