TSTP Solution File: ITP001+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : ITP001+1 : TPTP v8.1.2. Bugfixed v7.5.0.
% 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 : n018.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 04:15:35 EDT 2023
% Result : Theorem 0.21s 0.47s
% 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
% 0.00/0.12 % Problem : ITP001+1 : TPTP v8.1.2. Bugfixed v7.5.0.
% 0.12/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35 % Computer : n018.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Sun Aug 27 12:10:16 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.21/0.47 Command-line arguments: --no-flatten-goal
% 0.21/0.47
% 0.21/0.47 % SZS status Theorem
% 0.21/0.47
% 0.21/0.47 % SZS output start Proof
% 0.21/0.47 Take the following subset of the input axioms:
% 0.21/0.47 fof(reserved_2Eho_2Etruth, axiom, p(s(tyop_2Emin_2Ebool, c_2Ebool_2ET_2E0))).
% 0.21/0.47 fof(thm_2Ebool_2ETRUTH, conjecture, p(s(tyop_2Emin_2Ebool, c_2Ebool_2ET_2E0))).
% 0.21/0.47
% 0.21/0.47 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.47 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.47 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.47 fresh(y, y, x1...xn) = u
% 0.21/0.47 C => fresh(s, t, x1...xn) = v
% 0.21/0.47 where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.47 variables of u and v.
% 0.21/0.47 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.47 input problem has no model of domain size 1).
% 0.21/0.47
% 0.21/0.47 The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.47
% 0.21/0.47 Axiom 1 (reserved_2Eho_2Etruth): p(s(tyop_2Emin_2Ebool, c_2Ebool_2ET_2E0)) = true2.
% 0.21/0.47
% 0.21/0.47 Goal 1 (thm_2Ebool_2ETRUTH): p(s(tyop_2Emin_2Ebool, c_2Ebool_2ET_2E0)) = true2.
% 0.21/0.47 Proof:
% 0.21/0.47 p(s(tyop_2Emin_2Ebool, c_2Ebool_2ET_2E0))
% 0.21/0.47 = { by axiom 1 (reserved_2Eho_2Etruth) }
% 0.21/0.47 true2
% 0.21/0.47 % SZS output end Proof
% 0.21/0.47
% 0.21/0.47 RESULT: Theorem (the conjecture is true).
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