TSTP Solution File: NUN088+3 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUN088+3 : TPTP v8.1.2. Released v7.3.0.
% 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 : n012.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 12:51:56 EDT 2023
% Result : Theorem 0.20s 0.41s
% Output : Proof 0.20s
% 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 : NUN088+3 : TPTP v8.1.2. Released v7.3.0.
% 0.00/0.13 % 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 : n012.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 : Sun Aug 27 09:44:40 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.20/0.41 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.41
% 0.20/0.41 % SZS status Theorem
% 0.20/0.41
% 0.20/0.41 % SZS output start Proof
% 0.20/0.41 Take the following subset of the input axioms:
% 0.20/0.41 fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~r1(Y20) | Y20!=Y10) | ~r2(X7, Y10))).
% 0.20/0.41 fof(zerouneqone2, conjecture, ![Y0]: (![Y1]: (Y0!=Y1 | ~r2(Y0, Y1)) | ~r1(Y0))).
% 0.20/0.41
% 0.20/0.41 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.41 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.41 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.41 fresh(y, y, x1...xn) = u
% 0.20/0.41 C => fresh(s, t, x1...xn) = v
% 0.20/0.41 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.41 variables of u and v.
% 0.20/0.41 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.41 input problem has no model of domain size 1).
% 0.20/0.41
% 0.20/0.41 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.41
% 0.20/0.41 Axiom 1 (zerouneqone2): y0 = y1.
% 0.20/0.41 Axiom 2 (zerouneqone2_1): r1(y0) = true2.
% 0.20/0.41 Axiom 3 (zerouneqone2_2): r2(y0, y1) = true2.
% 0.20/0.41
% 0.20/0.41 Goal 1 (axiom_7a): tuple(r1(X), r2(Y, X)) = tuple(true2, true2).
% 0.20/0.41 The goal is true when:
% 0.20/0.41 X = y0
% 0.20/0.41 Y = y0
% 0.20/0.41
% 0.20/0.41 Proof:
% 0.20/0.41 tuple(r1(y0), r2(y0, y0))
% 0.20/0.41 = { by axiom 1 (zerouneqone2) }
% 0.20/0.41 tuple(r1(y0), r2(y0, y1))
% 0.20/0.41 = { by axiom 3 (zerouneqone2_2) }
% 0.20/0.41 tuple(r1(y0), true2)
% 0.20/0.41 = { by axiom 2 (zerouneqone2_1) }
% 0.20/0.41 tuple(true2, true2)
% 0.20/0.41 % SZS output end Proof
% 0.20/0.41
% 0.20/0.41 RESULT: Theorem (the conjecture is true).
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