TSTP Solution File: LCL460+1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : LCL460+1 : 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 : n002.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 08:19:05 EDT 2023

% Result   : Theorem 0.21s 0.49s
% Output   : Proof 0.21s
% 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  : LCL460+1 : TPTP v8.1.2. Released v3.3.0.
% 0.07/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35  % Computer : n002.cluster.edu
% 0.14/0.35  % Model    : x86_64 x86_64
% 0.14/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35  % Memory   : 8042.1875MB
% 0.14/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35  % CPULimit : 300
% 0.14/0.35  % WCLimit  : 300
% 0.14/0.35  % DateTime : Fri Aug 25 03:45:17 EDT 2023
% 0.14/0.35  % CPUTime  : 
% 0.21/0.49  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.21/0.49  
% 0.21/0.49  % SZS status Theorem
% 0.21/0.49  
% 0.21/0.49  % SZS output start Proof
% 0.21/0.49  Take the following subset of the input axioms:
% 0.21/0.49    fof(and_1, axiom, and_1 <=> ![X, Y]: is_a_theorem(implies(and(X, Y), X))).
% 0.21/0.49    fof(hilbert_and_1, axiom, and_1).
% 0.21/0.49    fof(kn2, axiom, kn2 <=> ![P, Q]: is_a_theorem(implies(and(P, Q), P))).
% 0.21/0.49    fof(rosser_kn2, conjecture, kn2).
% 0.21/0.49  
% 0.21/0.49  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.49  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.49  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.49    fresh(y, y, x1...xn) = u
% 0.21/0.49    C => fresh(s, t, x1...xn) = v
% 0.21/0.49  where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.49  variables of u and v.
% 0.21/0.49  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.49  input problem has no model of domain size 1).
% 0.21/0.49  
% 0.21/0.49  The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.49  
% 0.21/0.49  Axiom 1 (hilbert_and_1): and_1 = true.
% 0.21/0.49  Axiom 2 (kn2): fresh32(X, X) = true.
% 0.21/0.49  Axiom 3 (and_1_1): fresh58(X, X, Y, Z) = true.
% 0.21/0.49  Axiom 4 (and_1_1): fresh58(and_1, true, X, Y) = is_a_theorem(implies(and(X, Y), X)).
% 0.21/0.49  Axiom 5 (kn2): fresh32(is_a_theorem(implies(and(p10, q8), p10)), true) = kn2.
% 0.21/0.49  
% 0.21/0.49  Goal 1 (rosser_kn2): kn2 = true.
% 0.21/0.49  Proof:
% 0.21/0.49    kn2
% 0.21/0.49  = { by axiom 5 (kn2) R->L }
% 0.21/0.49    fresh32(is_a_theorem(implies(and(p10, q8), p10)), true)
% 0.21/0.49  = { by axiom 4 (and_1_1) R->L }
% 0.21/0.49    fresh32(fresh58(and_1, true, p10, q8), true)
% 0.21/0.49  = { by axiom 1 (hilbert_and_1) }
% 0.21/0.49    fresh32(fresh58(true, true, p10, q8), true)
% 0.21/0.49  = { by axiom 3 (and_1_1) }
% 0.21/0.49    fresh32(true, true)
% 0.21/0.49  = { by axiom 2 (kn2) }
% 0.21/0.49    true
% 0.21/0.49  % SZS output end Proof
% 0.21/0.49  
% 0.21/0.49  RESULT: Theorem (the conjecture is true).
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