TSTP Solution File: SET073+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SET073+1 : TPTP v8.1.2. Bugfixed v5.4.0.
% 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 : n019.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 15:31:03 EDT 2023
% Result : Theorem 0.18s 0.47s
% Output : Proof 0.18s
% 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 : SET073+1 : TPTP v8.1.2. Bugfixed v5.4.0.
% 0.00/0.12 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.32 % Computer : n019.cluster.edu
% 0.14/0.32 % Model : x86_64 x86_64
% 0.14/0.32 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.32 % Memory : 8042.1875MB
% 0.14/0.32 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.32 % CPULimit : 300
% 0.14/0.32 % WCLimit : 300
% 0.14/0.32 % DateTime : Sat Aug 26 16:23:43 EDT 2023
% 0.14/0.32 % CPUTime :
% 0.18/0.47 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.18/0.47
% 0.18/0.47 % SZS status Theorem
% 0.18/0.47
% 0.18/0.47 % SZS output start Proof
% 0.18/0.47 Take the following subset of the input axioms:
% 0.18/0.47 fof(complement, axiom, ![X, Z]: (member(Z, complement(X)) <=> (member(Z, universal_class) & ~member(Z, X)))).
% 0.18/0.47 fof(corollary1_1, conjecture, ![Y, X2]: (member(X2, universal_class) => unordered_pair(X2, Y)!=null_class)).
% 0.18/0.47 fof(disjoint_defn, axiom, ![X2, Y2]: (disjoint(X2, Y2) <=> ![U]: ~(member(U, X2) & member(U, Y2)))).
% 0.18/0.47 fof(domain_of, axiom, ![X2, Z2]: (member(Z2, domain_of(X2)) <=> (member(Z2, universal_class) & restrict(X2, singleton(Z2), universal_class)!=null_class))).
% 0.18/0.47 fof(null_class_defn, axiom, ![X2]: ~member(X2, null_class)).
% 0.18/0.47 fof(unordered_pair_defn, axiom, ![X2, Y2, U2]: (member(U2, unordered_pair(X2, Y2)) <=> (member(U2, universal_class) & (U2=X2 | U2=Y2)))).
% 0.18/0.47
% 0.18/0.47 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.18/0.47 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.18/0.47 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.18/0.47 fresh(y, y, x1...xn) = u
% 0.18/0.47 C => fresh(s, t, x1...xn) = v
% 0.18/0.47 where fresh is a fresh function symbol and x1..xn are the free
% 0.18/0.47 variables of u and v.
% 0.18/0.47 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.18/0.47 input problem has no model of domain size 1).
% 0.18/0.47
% 0.18/0.47 The encoding turns the above axioms into the following unit equations and goals:
% 0.18/0.47
% 0.18/0.47 Axiom 1 (corollary1_1_1): member(x, universal_class) = true2.
% 0.18/0.47 Axiom 2 (corollary1_1): unordered_pair(x, y) = null_class.
% 0.18/0.47 Axiom 3 (unordered_pair_defn_2): fresh9(X, X, Y, Z) = true2.
% 0.18/0.47 Axiom 4 (unordered_pair_defn_2): fresh9(member(X, universal_class), true2, X, Y) = member(X, unordered_pair(X, Y)).
% 0.18/0.47
% 0.18/0.47 Goal 1 (null_class_defn): member(X, null_class) = true2.
% 0.18/0.47 The goal is true when:
% 0.18/0.47 X = x
% 0.18/0.47
% 0.18/0.47 Proof:
% 0.18/0.47 member(x, null_class)
% 0.18/0.47 = { by axiom 2 (corollary1_1) R->L }
% 0.18/0.47 member(x, unordered_pair(x, y))
% 0.18/0.47 = { by axiom 4 (unordered_pair_defn_2) R->L }
% 0.18/0.47 fresh9(member(x, universal_class), true2, x, y)
% 0.18/0.47 = { by axiom 1 (corollary1_1_1) }
% 0.18/0.47 fresh9(true2, true2, x, y)
% 0.18/0.47 = { by axiom 3 (unordered_pair_defn_2) }
% 0.18/0.47 true2
% 0.18/0.47 % SZS output end Proof
% 0.18/0.47
% 0.18/0.47 RESULT: Theorem (the conjecture is true).
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