TSTP Solution File: KLE066+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : KLE066+1 : 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 : n015.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 05:35:45 EDT 2023
% Result : Theorem 0.20s 0.40s
% 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
% 0.00/0.13 % Problem : KLE066+1 : 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
% 0.14/0.35 % Computer : n015.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 300
% 0.14/0.35 % DateTime : Tue Aug 29 12:40:56 EDT 2023
% 0.14/0.35 % CPUTime :
% 0.20/0.40 Command-line arguments: --ground-connectedness --complete-subsets
% 0.20/0.40
% 0.20/0.40 % SZS status Theorem
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% 0.20/0.40 % SZS output start Proof
% 0.20/0.40 Take the following subset of the input axioms:
% 0.20/0.40 fof(additive_identity, axiom, ![A]: addition(A, zero)=A).
% 0.20/0.40 fof(domain1, axiom, ![X0]: addition(X0, multiplication(domain(X0), X0))=multiplication(domain(X0), X0)).
% 0.20/0.40 fof(domain2, axiom, ![X1, X0_2]: domain(multiplication(X0_2, X1))=domain(multiplication(X0_2, domain(X1)))).
% 0.20/0.40 fof(domain4, axiom, domain(zero)=zero).
% 0.20/0.40 fof(goals, conjecture, ![X0_2, X1_2]: (multiplication(X0_2, X1_2)=zero <= multiplication(X0_2, domain(X1_2))=zero)).
% 0.20/0.40 fof(left_annihilation, axiom, ![A2]: multiplication(zero, A2)=zero).
% 0.20/0.40
% 0.20/0.40 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.40 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.40 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.40 fresh(y, y, x1...xn) = u
% 0.20/0.40 C => fresh(s, t, x1...xn) = v
% 0.20/0.40 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.40 variables of u and v.
% 0.20/0.40 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.40 input problem has no model of domain size 1).
% 0.20/0.40
% 0.20/0.40 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.40
% 0.20/0.40 Axiom 1 (domain4): domain(zero) = zero.
% 0.20/0.40 Axiom 2 (additive_identity): addition(X, zero) = X.
% 0.20/0.40 Axiom 3 (left_annihilation): multiplication(zero, X) = zero.
% 0.20/0.40 Axiom 4 (goals): multiplication(x0, domain(x1)) = zero.
% 0.20/0.40 Axiom 5 (domain2): domain(multiplication(X, Y)) = domain(multiplication(X, domain(Y))).
% 0.20/0.40 Axiom 6 (domain1): addition(X, multiplication(domain(X), X)) = multiplication(domain(X), X).
% 0.20/0.40
% 0.20/0.40 Lemma 7: domain(multiplication(x0, x1)) = zero.
% 0.20/0.40 Proof:
% 0.20/0.40 domain(multiplication(x0, x1))
% 0.20/0.40 = { by axiom 5 (domain2) }
% 0.20/0.40 domain(multiplication(x0, domain(x1)))
% 0.20/0.40 = { by axiom 4 (goals) }
% 0.20/0.40 domain(zero)
% 0.20/0.40 = { by axiom 1 (domain4) }
% 0.20/0.40 zero
% 0.20/0.40
% 0.20/0.40 Goal 1 (goals_1): multiplication(x0, x1) = zero.
% 0.20/0.40 Proof:
% 0.20/0.40 multiplication(x0, x1)
% 0.20/0.40 = { by axiom 2 (additive_identity) R->L }
% 0.20/0.40 addition(multiplication(x0, x1), zero)
% 0.20/0.40 = { by axiom 3 (left_annihilation) R->L }
% 0.20/0.40 addition(multiplication(x0, x1), multiplication(zero, multiplication(x0, x1)))
% 0.20/0.40 = { by lemma 7 R->L }
% 0.20/0.40 addition(multiplication(x0, x1), multiplication(domain(multiplication(x0, x1)), multiplication(x0, x1)))
% 0.20/0.40 = { by axiom 6 (domain1) }
% 0.20/0.40 multiplication(domain(multiplication(x0, x1)), multiplication(x0, x1))
% 0.20/0.40 = { by lemma 7 }
% 0.20/0.40 multiplication(zero, multiplication(x0, x1))
% 0.20/0.40 = { by axiom 3 (left_annihilation) }
% 0.20/0.40 zero
% 0.20/0.40 % SZS output end Proof
% 0.20/0.40
% 0.20/0.40 RESULT: Theorem (the conjecture is true).
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