TSTP Solution File: NUM565+3 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM565+3 : 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 : n029.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 11:57:01 EDT 2023
% Result : Theorem 3.68s 0.93s
% Output : Proof 3.68s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : NUM565+3 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.18/0.35 % Computer : n029.cluster.edu
% 0.18/0.35 % Model : x86_64 x86_64
% 0.18/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.18/0.35 % Memory : 8042.1875MB
% 0.18/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.18/0.35 % CPULimit : 300
% 0.18/0.35 % WCLimit : 300
% 0.18/0.35 % DateTime : Fri Aug 25 18:18:39 EDT 2023
% 0.18/0.35 % CPUTime :
% 3.68/0.93 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 3.68/0.93
% 3.68/0.93 % SZS status Theorem
% 3.68/0.93
% 3.68/0.93 % SZS output start Proof
% 3.68/0.93 Take the following subset of the input axioms:
% 3.68/0.93 fof(mCardEmpty, axiom, ![W0]: (aSet0(W0) => (sbrdtbr0(W0)=sz00 <=> W0=slcrc0))).
% 3.68/0.93 fof(m__, conjecture, ![W0_2]: ((aSet0(W0_2) & (![W1]: (aElementOf0(W1, W0_2) => aElementOf0(W1, xS)) & (aSubsetOf0(W0_2, xS) & (sbrdtbr0(W0_2)=sz00 & aElementOf0(W0_2, slbdtsldtrb0(xS, sz00)))))) => ((aSet0(slcrc0) & ~?[W1_2]: aElementOf0(W1_2, slcrc0)) => sdtlpdtrp0(xc, W0_2)=sdtlpdtrp0(xc, slcrc0)))).
% 3.68/0.93
% 3.68/0.93 Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.68/0.93 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.68/0.93 We repeatedly replace C & s=t => u=v by the two clauses:
% 3.68/0.93 fresh(y, y, x1...xn) = u
% 3.68/0.93 C => fresh(s, t, x1...xn) = v
% 3.68/0.93 where fresh is a fresh function symbol and x1..xn are the free
% 3.68/0.93 variables of u and v.
% 3.68/0.93 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.68/0.93 input problem has no model of domain size 1).
% 3.68/0.93
% 3.68/0.93 The encoding turns the above axioms into the following unit equations and goals:
% 3.68/0.93
% 3.68/0.93 Axiom 1 (m___2): aSet0(w0) = true2.
% 3.68/0.93 Axiom 2 (m__): sbrdtbr0(w0) = sz00.
% 3.68/0.93 Axiom 3 (mCardEmpty_1): fresh126(X, X, Y) = slcrc0.
% 3.68/0.93 Axiom 4 (mCardEmpty_1): fresh7(X, X, Y) = Y.
% 3.68/0.93 Axiom 5 (mCardEmpty_1): fresh7(aSet0(X), true2, X) = fresh126(sbrdtbr0(X), sz00, X).
% 3.68/0.93
% 3.68/0.93 Goal 1 (m___5): sdtlpdtrp0(xc, w0) = sdtlpdtrp0(xc, slcrc0).
% 3.68/0.93 Proof:
% 3.68/0.93 sdtlpdtrp0(xc, w0)
% 3.68/0.93 = { by axiom 4 (mCardEmpty_1) R->L }
% 3.68/0.93 sdtlpdtrp0(xc, fresh7(true2, true2, w0))
% 3.68/0.93 = { by axiom 1 (m___2) R->L }
% 3.68/0.93 sdtlpdtrp0(xc, fresh7(aSet0(w0), true2, w0))
% 3.68/0.93 = { by axiom 5 (mCardEmpty_1) }
% 3.68/0.93 sdtlpdtrp0(xc, fresh126(sbrdtbr0(w0), sz00, w0))
% 3.68/0.93 = { by axiom 2 (m__) }
% 3.68/0.93 sdtlpdtrp0(xc, fresh126(sz00, sz00, w0))
% 3.68/0.93 = { by axiom 3 (mCardEmpty_1) }
% 3.68/0.93 sdtlpdtrp0(xc, slcrc0)
% 3.68/0.93 % SZS output end Proof
% 3.68/0.93
% 3.68/0.93 RESULT: Theorem (the conjecture is true).
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