TSTP Solution File: LAT388+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : LAT388+1 : 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 : n004.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 06:29:01 EDT 2023
% Result : Theorem 0.19s 0.50s
% Output : Proof 0.19s
% 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 : LAT388+1 : 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
% 0.13/0.33 % Computer : n004.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % 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 : Thu Aug 24 04:43:08 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.50 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.19/0.50
% 0.19/0.50 % SZS status Theorem
% 0.19/0.50
% 0.19/0.50 % SZS output start Proof
% 0.19/0.50 Take the following subset of the input axioms:
% 0.19/0.50 fof(mDefEmpty, definition, ![W0]: (aSet0(W0) => (isEmpty0(W0) <=> ~?[W1]: aElementOf0(W1, W0)))).
% 0.19/0.50 fof(m__, conjecture, ?[W0_2]: aSupremumOfIn0(W0_2, xT, xS)).
% 0.19/0.50 fof(m__1330, hypothesis, aFixedPointOf0(xp, xf) & aSupremumOfIn0(xp, xT, xS)).
% 0.19/0.50
% 0.19/0.50 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.50 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.50 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.50 fresh(y, y, x1...xn) = u
% 0.19/0.50 C => fresh(s, t, x1...xn) = v
% 0.19/0.50 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.50 variables of u and v.
% 0.19/0.50 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.50 input problem has no model of domain size 1).
% 0.19/0.50
% 0.19/0.50 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.50
% 0.19/0.50 Axiom 1 (m__1330): aSupremumOfIn0(xp, xT, xS) = true2.
% 0.19/0.50
% 0.19/0.50 Goal 1 (m__): aSupremumOfIn0(X, xT, xS) = true2.
% 0.19/0.50 The goal is true when:
% 0.19/0.50 X = xp
% 0.19/0.50
% 0.19/0.50 Proof:
% 0.19/0.50 aSupremumOfIn0(xp, xT, xS)
% 0.19/0.50 = { by axiom 1 (m__1330) }
% 0.19/0.50 true2
% 0.19/0.50 % SZS output end Proof
% 0.19/0.50
% 0.19/0.50 RESULT: Theorem (the conjecture is true).
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