%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN000+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 : n023.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 03:32:39 EDT 2023

% Result   : Theorem 0.19s 0.40s
% Output   : Proof 0.19s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12  % Problem  : SYN000+1 : TPTP v8.1.2. Released v4.0.0.
% 0.06/0.13  % 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 : n023.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 : Sat Aug 26 21:15:56 EDT 2023
% 0.13/0.34  % CPUTime  : 
% 0.19/0.40  Command-line arguments: --no-flatten-goal
% 0.19/0.40  
% 0.19/0.40  % SZS status Theorem
% 0.19/0.40  
% 0.19/0.40  % SZS output start Proof
% 0.19/0.40  Take the following subset of the input axioms:
% 0.19/0.40    fof('123', axiom, ![X]: ((p(X) | ~q(X, a)) => ?[Y, Z]: (r(X, f(Y), g(X, f(Y), Z)) & ~s(f(f(f(b))))))).
% 0.19/0.40    fof(first_order, axiom, ![X2]: ((p(X2) | ~q(X2, a)) => ?[Y2, Z2]: (r(X2, f(Y2), g(X2, f(Y2), Z2)) & ~s(f(f(f(b))))))).
% 0.19/0.40    fof(ia1, axiom, ia1).
% 0.19/0.40    fof(role_conjecture, conjecture, ?[X2]: p(X2)).
% 0.19/0.40    fof(role_hypothesis, hypothesis, p(h)).
% 0.19/0.40  
% 0.19/0.40  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.40  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.40  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.40    fresh(y, y, x1...xn) = u
% 0.19/0.40    C => fresh(s, t, x1...xn) = v
% 0.19/0.40  where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.40  variables of u and v.
% 0.19/0.40  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.40  input problem has no model of domain size 1).
% 0.19/0.40  
% 0.19/0.40  The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.40  
% 0.19/0.40  Axiom 1 (ia1): ia1 = true2.
% 0.19/0.40  Axiom 2 (role_hypothesis): p(h) = true2.
% 0.19/0.40  
% 0.19/0.40  Goal 1 (role_conjecture): p(X) = true2.
% 0.19/0.40  The goal is true when:
% 0.19/0.40    X = h
% 0.19/0.40  
% 0.19/0.40  Proof:
% 0.19/0.40    p(h)
% 0.19/0.40  = { by axiom 2 (role_hypothesis) }
% 0.19/0.40    true2
% 0.19/0.40  = { by axiom 1 (ia1) R->L }
% 0.19/0.40    ia1
% 0.19/0.40  = { by axiom 1 (ia1) }
% 0.19/0.40    true2
% 0.19/0.40  % SZS output end Proof
% 0.19/0.40  
% 0.19/0.40  RESULT: Theorem (the conjecture is true).
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