TSTP Solution File: KLE150+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : KLE150+2 : 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 : n003.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 05:36:05 EDT 2023
% Result : Theorem 0.19s 0.39s
% Output : Proof 0.19s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12 % Problem : KLE150+2 : TPTP v8.1.2. Released v4.0.0.
% 0.06/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34 % Computer : n003.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Tue Aug 29 11:14:55 EDT 2023
% 0.12/0.34 % CPUTime :
% 0.19/0.39 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.19/0.39
% 0.19/0.39 % SZS status Theorem
% 0.19/0.39
% 0.19/0.40 % SZS output start Proof
% 0.19/0.40 Take the following subset of the input axioms:
% 0.19/0.40 fof(additive_commutativity, axiom, ![A, B]: addition(A, B)=addition(B, A)).
% 0.19/0.40 fof(goals, conjecture, ![X0]: (leq(strong_iteration(multiplication(X0, zero)), addition(one, multiplication(X0, zero))) & leq(addition(one, multiplication(X0, zero)), strong_iteration(multiplication(X0, zero))))).
% 0.19/0.40 fof(idempotence, axiom, ![A3]: addition(A3, A3)=A3).
% 0.19/0.40 fof(infty_unfold1, axiom, ![A3]: strong_iteration(A3)=addition(multiplication(A3, strong_iteration(A3)), one)).
% 0.19/0.40 fof(left_annihilation, axiom, ![A3]: multiplication(zero, A3)=zero).
% 0.19/0.40 fof(multiplicative_associativity, axiom, ![C, B2, A3]: multiplication(A3, multiplication(B2, C))=multiplication(multiplication(A3, B2), C)).
% 0.19/0.40 fof(order, axiom, ![A2, B2]: (leq(A2, B2) <=> addition(A2, B2)=B2)).
% 0.19/0.40
% 0.19/0.40 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.40 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.40 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.40 fresh(y, y, x1...xn) = u
% 0.19/0.40 C => fresh(s, t, x1...xn) = v
% 0.19/0.40 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.40 variables of u and v.
% 0.19/0.40 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.40 input problem has no model of domain size 1).
% 0.19/0.40
% 0.19/0.40 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.40
% 0.19/0.40 Axiom 1 (left_annihilation): multiplication(zero, X) = zero.
% 0.19/0.40 Axiom 2 (idempotence): addition(X, X) = X.
% 0.19/0.40 Axiom 3 (additive_commutativity): addition(X, Y) = addition(Y, X).
% 0.19/0.40 Axiom 4 (multiplicative_associativity): multiplication(X, multiplication(Y, Z)) = multiplication(multiplication(X, Y), Z).
% 0.19/0.40 Axiom 5 (infty_unfold1): strong_iteration(X) = addition(multiplication(X, strong_iteration(X)), one).
% 0.19/0.40 Axiom 6 (order): fresh5(X, X, Y, Z) = true.
% 0.19/0.40 Axiom 7 (order): fresh5(addition(X, Y), Y, X, Y) = leq(X, Y).
% 0.19/0.40
% 0.19/0.40 Lemma 8: leq(X, X) = true.
% 0.19/0.40 Proof:
% 0.19/0.40 leq(X, X)
% 0.19/0.40 = { by axiom 7 (order) R->L }
% 0.19/0.40 fresh5(addition(X, X), X, X, X)
% 0.19/0.40 = { by axiom 2 (idempotence) }
% 0.19/0.40 fresh5(X, X, X, X)
% 0.19/0.40 = { by axiom 6 (order) }
% 0.19/0.40 true
% 0.19/0.40
% 0.19/0.40 Lemma 9: strong_iteration(multiplication(X, zero)) = addition(one, multiplication(X, zero)).
% 0.19/0.40 Proof:
% 0.19/0.40 strong_iteration(multiplication(X, zero))
% 0.19/0.40 = { by axiom 5 (infty_unfold1) }
% 0.19/0.40 addition(multiplication(multiplication(X, zero), strong_iteration(multiplication(X, zero))), one)
% 0.19/0.40 = { by axiom 3 (additive_commutativity) }
% 0.19/0.40 addition(one, multiplication(multiplication(X, zero), strong_iteration(multiplication(X, zero))))
% 0.19/0.40 = { by axiom 4 (multiplicative_associativity) R->L }
% 0.19/0.40 addition(one, multiplication(X, multiplication(zero, strong_iteration(multiplication(X, zero)))))
% 0.19/0.40 = { by axiom 1 (left_annihilation) }
% 0.19/0.40 addition(one, multiplication(X, zero))
% 0.19/0.40
% 0.19/0.40 Goal 1 (goals): tuple(leq(addition(one, multiplication(x0, zero)), strong_iteration(multiplication(x0, zero))), leq(strong_iteration(multiplication(x0_2, zero)), addition(one, multiplication(x0_2, zero)))) = tuple(true, true).
% 0.19/0.40 Proof:
% 0.19/0.40 tuple(leq(addition(one, multiplication(x0, zero)), strong_iteration(multiplication(x0, zero))), leq(strong_iteration(multiplication(x0_2, zero)), addition(one, multiplication(x0_2, zero))))
% 0.19/0.40 = { by lemma 9 }
% 0.19/0.40 tuple(leq(addition(one, multiplication(x0, zero)), addition(one, multiplication(x0, zero))), leq(strong_iteration(multiplication(x0_2, zero)), addition(one, multiplication(x0_2, zero))))
% 0.19/0.40 = { by lemma 8 }
% 0.19/0.40 tuple(true, leq(strong_iteration(multiplication(x0_2, zero)), addition(one, multiplication(x0_2, zero))))
% 0.19/0.40 = { by lemma 9 }
% 0.19/0.40 tuple(true, leq(addition(one, multiplication(x0_2, zero)), addition(one, multiplication(x0_2, zero))))
% 0.19/0.40 = { by lemma 8 }
% 0.19/0.40 tuple(true, true)
% 0.19/0.40 % SZS output end Proof
% 0.19/0.40
% 0.19/0.40 RESULT: Theorem (the conjecture is true).
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