TSTP Solution File: LCL664+1.001 by Twee---2.4.2
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- Process Solution
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% File : Twee---2.4.2
% Problem : LCL664+1.001 : 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 : n006.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 08:19:56 EDT 2023
% Result : Theorem 0.21s 0.40s
% 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 : LCL664+1.001 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35 % Computer : n006.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 05:02:07 EDT 2023
% 0.14/0.35 % CPUTime :
% 0.21/0.40 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.21/0.40
% 0.21/0.40 % SZS status Theorem
% 0.21/0.40
% 0.21/0.40 % SZS output start Proof
% 0.21/0.40 Take the following subset of the input axioms:
% 0.21/0.40 fof(main, conjecture, ~?[X]: ~(~![Y]: (~r1(X, Y) | p16(Y)) | (~![Y2]: (~r1(X, Y2) | p12(Y2)) | (~![Y2]: (~r1(X, Y2) | p14(Y2)) | (~![Y2]: (~r1(X, Y2) | p12(Y2)) | (![Y2]: (~r1(X, Y2) | p15(Y2)) | (![Y2]: (~r1(X, Y2) | p13(Y2)) | (![Y2]: (~r1(X, Y2) | p12(Y2)) | ![Y2]: (~r1(X, Y2) | p11(Y2)))))))))).
% 0.21/0.40
% 0.21/0.40 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.40 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.40 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.40 fresh(y, y, x1...xn) = u
% 0.21/0.40 C => fresh(s, t, x1...xn) = v
% 0.21/0.40 where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.40 variables of u and v.
% 0.21/0.40 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.40 input problem has no model of domain size 1).
% 0.21/0.40
% 0.21/0.40 The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.40
% 0.21/0.40 Axiom 1 (main_2): r1(x, y2) = true.
% 0.21/0.40 Axiom 2 (main_7): fresh(X, X, Y) = true.
% 0.21/0.40 Axiom 3 (main_7): fresh(r1(x, X), true, X) = p12(X).
% 0.21/0.40
% 0.21/0.40 Goal 1 (main_8): p12(y2) = true.
% 0.21/0.40 Proof:
% 0.21/0.40 p12(y2)
% 0.21/0.40 = { by axiom 3 (main_7) R->L }
% 0.21/0.40 fresh(r1(x, y2), true, y2)
% 0.21/0.40 = { by axiom 1 (main_2) }
% 0.21/0.40 fresh(true, true, y2)
% 0.21/0.40 = { by axiom 2 (main_7) }
% 0.21/0.40 true
% 0.21/0.40 % SZS output end Proof
% 0.21/0.40
% 0.21/0.40 RESULT: Theorem (the conjecture is true).
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