TSTP Solution File: LAT388+4 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : LAT388+4 : 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 : n032.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 0.16s 0.54s
% Output : Proof 0.16s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.10 % Problem : LAT388+4 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.11 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.11/0.30 % Computer : n032.cluster.edu
% 0.11/0.30 % Model : x86_64 x86_64
% 0.11/0.30 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.30 % Memory : 8042.1875MB
% 0.11/0.30 % OS : Linux 3.10.0-693.el7.x86_64
% 0.11/0.30 % CPULimit : 300
% 0.11/0.30 % WCLimit : 300
% 0.11/0.30 % DateTime : Thu Aug 24 06:32:41 EDT 2023
% 0.11/0.30 % CPUTime :
% 0.16/0.54 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.16/0.54
% 0.16/0.54 % SZS status Theorem
% 0.16/0.54
% 0.16/0.54 % SZS output start Proof
% 0.16/0.54 Take the following subset of the input axioms:
% 0.16/0.54 fof(mDefEmpty, definition, ![W0]: (aSet0(W0) => (isEmpty0(W0) <=> ~?[W1]: aElementOf0(W1, W0)))).
% 0.16/0.54 fof(m__, conjecture, ?[W0_2]: ((aElementOf0(W0_2, xS) & (((aElementOf0(W0_2, xS) & ![W1_2]: (aElementOf0(W1_2, xT) => sdtlseqdt0(W1_2, W0_2))) | aUpperBoundOfIn0(W0_2, xT, xS)) & ![W1_2]: ((aElementOf0(W1_2, xS) & (![W2]: (aElementOf0(W2, xT) => sdtlseqdt0(W2, W1_2)) & aUpperBoundOfIn0(W1_2, xT, xS))) => sdtlseqdt0(W0_2, W1_2)))) | aSupremumOfIn0(W0_2, xT, xS))).
% 0.16/0.54 fof(m__1330, hypothesis, aElementOf0(xp, szDzozmdt0(xf)) & (sdtlpdtrp0(xf, xp)=xp & (aFixedPointOf0(xp, xf) & (![W0_2]: (aElementOf0(W0_2, xT) => sdtlseqdt0(W0_2, xp)) & (aUpperBoundOfIn0(xp, xT, xS) & (![W0_2]: (((aElementOf0(W0_2, xS) & ![W1_2]: (aElementOf0(W1_2, xT) => sdtlseqdt0(W1_2, W0_2))) | aUpperBoundOfIn0(W0_2, xT, xS)) => sdtlseqdt0(xp, W0_2)) & aSupremumOfIn0(xp, xT, xS))))))).
% 0.16/0.54
% 0.16/0.54 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.16/0.54 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.16/0.54 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.16/0.54 fresh(y, y, x1...xn) = u
% 0.16/0.54 C => fresh(s, t, x1...xn) = v
% 0.16/0.54 where fresh is a fresh function symbol and x1..xn are the free
% 0.16/0.54 variables of u and v.
% 0.16/0.54 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.16/0.54 input problem has no model of domain size 1).
% 0.16/0.54
% 0.16/0.54 The encoding turns the above axioms into the following unit equations and goals:
% 0.16/0.54
% 0.16/0.54 Axiom 1 (m__1330_3): aSupremumOfIn0(xp, xT, xS) = true2.
% 0.16/0.54
% 0.16/0.54 Goal 1 (m___5): aSupremumOfIn0(X, xT, xS) = true2.
% 0.16/0.54 The goal is true when:
% 0.16/0.54 X = xp
% 0.16/0.54
% 0.16/0.54 Proof:
% 0.16/0.54 aSupremumOfIn0(xp, xT, xS)
% 0.16/0.54 = { by axiom 1 (m__1330_3) }
% 0.16/0.54 true2
% 0.16/0.54 % SZS output end Proof
% 0.16/0.54
% 0.16/0.54 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------