TSTP Solution File: SYN721+1 by Twee---2.4.1

View Problem - Process Solution

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% File     : Twee---2.4.1
% Problem  : SYN721+1 : TPTP v8.1.0. Released v2.5.0.
% 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 : n017.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 Sep 29 23:25:15 EDT 2022

% Result   : Theorem 0.13s 0.38s
% Output   : Proof 0.13s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem  : SYN721+1 : TPTP v8.1.0. Released v2.5.0.
% 0.07/0.13  % 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 : n017.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 : Mon Sep  5 06:09:25 EDT 2022
% 0.13/0.34  % CPUTime  : 
% 0.13/0.38  % SZS status Theorem
% 0.13/0.38  
% 0.13/0.38  % SZS output start Proof
% 0.13/0.38  Take the following subset of the input axioms:
% 0.13/0.38    fof(lx1, conjecture, (r(a, b) & (![X]: (?[Y]: r(X, Y) => q(X, X)) & ![U, V]: (q(U, V) => ![Z]: r(Z, V)))) => ?[W]: (r(b, W) & q(W, a))).
% 0.13/0.38  
% 0.13/0.38  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.13/0.38  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.13/0.38  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.13/0.38    fresh(y, y, x1...xn) = u
% 0.13/0.38    C => fresh(s, t, x1...xn) = v
% 0.13/0.38  where fresh is a fresh function symbol and x1..xn are the free
% 0.13/0.38  variables of u and v.
% 0.13/0.38  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.13/0.38  input problem has no model of domain size 1).
% 0.13/0.38  
% 0.13/0.38  The encoding turns the above axioms into the following unit equations and goals:
% 0.13/0.38  
% 0.13/0.38  Axiom 1 (lx1): r(a, b) = true2.
% 0.13/0.38  Axiom 2 (lx1_1): fresh(X, X, Y) = true2.
% 0.13/0.38  Axiom 3 (lx1_3): fresh2(X, X, Y, Z) = true2.
% 0.13/0.38  Axiom 4 (lx1_1): fresh(r(X, Y), true2, X) = q(X, X).
% 0.13/0.38  Axiom 5 (lx1_3): fresh2(q(X, Y), true2, Y, Z) = r(Z, Y).
% 0.13/0.38  
% 0.13/0.38  Lemma 6: q(a, a) = true2.
% 0.13/0.38  Proof:
% 0.13/0.38    q(a, a)
% 0.13/0.38  = { by axiom 4 (lx1_1) R->L }
% 0.13/0.38    fresh(r(a, b), true2, a)
% 0.13/0.38  = { by axiom 1 (lx1) }
% 0.13/0.38    fresh(true2, true2, a)
% 0.13/0.38  = { by axiom 2 (lx1_1) }
% 0.13/0.38    true2
% 0.13/0.38  
% 0.13/0.38  Goal 1 (lx1_2): tuple(r(b, X), q(X, a)) = tuple(true2, true2).
% 0.13/0.38  The goal is true when:
% 0.13/0.38    X = a
% 0.13/0.38  
% 0.13/0.38  Proof:
% 0.13/0.39    tuple(r(b, a), q(a, a))
% 0.13/0.39  = { by axiom 5 (lx1_3) R->L }
% 0.13/0.39    tuple(fresh2(q(a, a), true2, a, b), q(a, a))
% 0.13/0.39  = { by lemma 6 }
% 0.13/0.39    tuple(fresh2(true2, true2, a, b), q(a, a))
% 0.13/0.39  = { by axiom 3 (lx1_3) }
% 0.13/0.39    tuple(true2, q(a, a))
% 0.13/0.39  = { by lemma 6 }
% 0.13/0.39    tuple(true2, true2)
% 0.13/0.39  % SZS output end Proof
% 0.13/0.39  
% 0.13/0.39  RESULT: Theorem (the conjecture is true).
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