TSTP Solution File: SET881+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SET881+1 : TPTP v8.1.2. Released v3.2.0.
% 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 : n008.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 15:33:38 EDT 2023
% Result : Theorem 0.13s 0.39s
% 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 : SET881+1 : TPTP v8.1.2. Released v3.2.0.
% 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 : n008.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 : Sat Aug 26 12:10:47 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.13/0.39 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.13/0.39
% 0.13/0.39 % SZS status Theorem
% 0.13/0.39
% 0.13/0.39 % SZS output start Proof
% 0.13/0.39 Take the following subset of the input axioms:
% 0.20/0.39 fof(d2_tarski, axiom, ![B, C, A2]: (C=unordered_pair(A2, B) <=> ![D]: (in(D, C) <=> (D=A2 | D=B)))).
% 0.20/0.39 fof(l36_zfmisc_1, axiom, ![B2, A2_2]: (set_difference(singleton(A2_2), B2)=empty_set <=> in(A2_2, B2))).
% 0.20/0.39 fof(t22_zfmisc_1, conjecture, ![A, B2]: set_difference(singleton(A), unordered_pair(A, B2))=empty_set).
% 0.20/0.39
% 0.20/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.39 fresh(y, y, x1...xn) = u
% 0.20/0.39 C => fresh(s, t, x1...xn) = v
% 0.20/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.39 variables of u and v.
% 0.20/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.39 input problem has no model of domain size 1).
% 0.20/0.39
% 0.20/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.39
% 0.20/0.39 Axiom 1 (d2_tarski_1): equiv(X, Y, X) = true2.
% 0.20/0.39 Axiom 2 (d2_tarski_4): fresh5(X, X, Y, Z) = true2.
% 0.20/0.39 Axiom 3 (l36_zfmisc_1_1): fresh2(X, X, Y, Z) = empty_set.
% 0.20/0.39 Axiom 4 (d2_tarski_4): fresh6(X, X, Y, Z, W, V) = in(V, W).
% 0.20/0.39 Axiom 5 (l36_zfmisc_1_1): fresh2(in(X, Y), true2, X, Y) = set_difference(singleton(X), Y).
% 0.20/0.39 Axiom 6 (d2_tarski_4): fresh6(equiv(X, Y, Z), true2, X, Y, W, Z) = fresh5(W, unordered_pair(X, Y), W, Z).
% 0.20/0.39
% 0.20/0.39 Goal 1 (t22_zfmisc_1): set_difference(singleton(a), unordered_pair(a, b)) = empty_set.
% 0.20/0.39 Proof:
% 0.20/0.39 set_difference(singleton(a), unordered_pair(a, b))
% 0.20/0.39 = { by axiom 5 (l36_zfmisc_1_1) R->L }
% 0.20/0.39 fresh2(in(a, unordered_pair(a, b)), true2, a, unordered_pair(a, b))
% 0.20/0.39 = { by axiom 4 (d2_tarski_4) R->L }
% 0.20/0.39 fresh2(fresh6(true2, true2, a, b, unordered_pair(a, b), a), true2, a, unordered_pair(a, b))
% 0.20/0.39 = { by axiom 1 (d2_tarski_1) R->L }
% 0.20/0.39 fresh2(fresh6(equiv(a, b, a), true2, a, b, unordered_pair(a, b), a), true2, a, unordered_pair(a, b))
% 0.20/0.39 = { by axiom 6 (d2_tarski_4) }
% 0.20/0.39 fresh2(fresh5(unordered_pair(a, b), unordered_pair(a, b), unordered_pair(a, b), a), true2, a, unordered_pair(a, b))
% 0.20/0.39 = { by axiom 2 (d2_tarski_4) }
% 0.20/0.39 fresh2(true2, true2, a, unordered_pair(a, b))
% 0.20/0.39 = { by axiom 3 (l36_zfmisc_1_1) }
% 0.20/0.39 empty_set
% 0.20/0.39 % SZS output end Proof
% 0.20/0.39
% 0.20/0.39 RESULT: Theorem (the conjecture is true).
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