TSTP Solution File: SEU294+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SEU294+2 : TPTP v8.1.2. Released v3.3.0.
% 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 : n028.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 17:52:03 EDT 2023
% Result : Theorem 50.40s 6.88s
% Output : Proof 50.40s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SEU294+2 : TPTP v8.1.2. Released v3.3.0.
% 0.07/0.13 % 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 : n028.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 : Thu Aug 24 01:37:19 EDT 2023
% 0.13/0.34 % CPUTime :
% 50.40/6.88 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 50.40/6.88
% 50.40/6.88 % SZS status Theorem
% 50.40/6.88
% 50.40/6.88 % SZS output start Proof
% 50.40/6.88 Take the following subset of the input axioms:
% 50.40/6.88 fof(cc2_finset_1, axiom, ![A2]: (finite(A2) => ![B]: (element(B, powerset(A2)) => finite(B)))).
% 50.40/6.88 fof(t13_finset_1, conjecture, ![A, B2]: ((subset(A, B2) & finite(B2)) => finite(A))).
% 50.40/6.88 fof(t3_subset, axiom, ![B2, A2_2]: (element(A2_2, powerset(B2)) <=> subset(A2_2, B2))).
% 50.40/6.88
% 50.40/6.88 Now clausify the problem and encode Horn clauses using encoding 3 of
% 50.40/6.88 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 50.40/6.88 We repeatedly replace C & s=t => u=v by the two clauses:
% 50.40/6.88 fresh(y, y, x1...xn) = u
% 50.40/6.88 C => fresh(s, t, x1...xn) = v
% 50.40/6.88 where fresh is a fresh function symbol and x1..xn are the free
% 50.40/6.88 variables of u and v.
% 50.40/6.88 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 50.40/6.88 input problem has no model of domain size 1).
% 50.40/6.88
% 50.40/6.88 The encoding turns the above axioms into the following unit equations and goals:
% 50.40/6.88
% 50.40/6.88 Axiom 1 (t13_finset_1): finite(b6) = true2.
% 50.40/6.88 Axiom 2 (t13_finset_1_1): subset(a, b6) = true2.
% 50.40/6.88 Axiom 3 (cc2_finset_1): fresh726(X, X, Y) = true2.
% 50.40/6.88 Axiom 4 (cc2_finset_1): fresh727(X, X, Y, Z) = finite(Z).
% 50.40/6.88 Axiom 5 (t3_subset_1): fresh147(X, X, Y, Z) = true2.
% 50.40/6.88 Axiom 6 (cc2_finset_1): fresh727(finite(X), true2, X, Y) = fresh726(element(Y, powerset(X)), true2, Y).
% 50.40/6.88 Axiom 7 (t3_subset_1): fresh147(subset(X, Y), true2, X, Y) = element(X, powerset(Y)).
% 50.40/6.88
% 50.40/6.88 Goal 1 (t13_finset_1_2): finite(a) = true2.
% 50.40/6.88 Proof:
% 50.40/6.88 finite(a)
% 50.40/6.88 = { by axiom 4 (cc2_finset_1) R->L }
% 50.40/6.88 fresh727(true2, true2, b6, a)
% 50.40/6.88 = { by axiom 1 (t13_finset_1) R->L }
% 50.40/6.88 fresh727(finite(b6), true2, b6, a)
% 50.40/6.88 = { by axiom 6 (cc2_finset_1) }
% 50.40/6.88 fresh726(element(a, powerset(b6)), true2, a)
% 50.40/6.88 = { by axiom 7 (t3_subset_1) R->L }
% 50.40/6.88 fresh726(fresh147(subset(a, b6), true2, a, b6), true2, a)
% 50.40/6.88 = { by axiom 2 (t13_finset_1_1) }
% 50.40/6.88 fresh726(fresh147(true2, true2, a, b6), true2, a)
% 50.40/6.88 = { by axiom 5 (t3_subset_1) }
% 50.40/6.88 fresh726(true2, true2, a)
% 50.40/6.88 = { by axiom 3 (cc2_finset_1) }
% 50.40/6.88 true2
% 50.40/6.88 % SZS output end Proof
% 50.40/6.88
% 50.40/6.88 RESULT: Theorem (the conjecture is true).
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