TSTP Solution File: KLE053+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : KLE053+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 : n004.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 05:35:42 EDT 2023
% Result : Theorem 0.20s 0.38s
% 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
% 0.12/0.12 % Problem : KLE053+1 : TPTP v8.1.2. Released v4.0.0.
% 0.12/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.34 % Computer : n004.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Tue Aug 29 11:44:07 EDT 2023
% 0.14/0.35 % CPUTime :
% 0.20/0.38 Command-line arguments: --flatten
% 0.20/0.38
% 0.20/0.38 % SZS status Theorem
% 0.20/0.38
% 0.20/0.38 % SZS output start Proof
% 0.20/0.38 Take the following subset of the input axioms:
% 0.20/0.38 fof(domain2, axiom, ![X0, X1]: domain(multiplication(X0, X1))=domain(multiplication(X0, domain(X1)))).
% 0.20/0.38 fof(goals, conjecture, ![X0_2]: domain(domain(X0_2))=domain(X0_2)).
% 0.20/0.38 fof(multiplicative_left_identity, axiom, ![A]: multiplication(one, A)=A).
% 0.20/0.38
% 0.20/0.38 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.38 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.38 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.38 fresh(y, y, x1...xn) = u
% 0.20/0.38 C => fresh(s, t, x1...xn) = v
% 0.20/0.38 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.38 variables of u and v.
% 0.20/0.38 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.38 input problem has no model of domain size 1).
% 0.20/0.38
% 0.20/0.38 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.38
% 0.20/0.38 Axiom 1 (multiplicative_left_identity): multiplication(one, X) = X.
% 0.20/0.39 Axiom 2 (domain2): domain(multiplication(X, Y)) = domain(multiplication(X, domain(Y))).
% 0.20/0.39
% 0.20/0.39 Goal 1 (goals): domain(domain(x0)) = domain(x0).
% 0.20/0.39 Proof:
% 0.20/0.39 domain(domain(x0))
% 0.20/0.39 = { by axiom 1 (multiplicative_left_identity) R->L }
% 0.20/0.39 domain(multiplication(one, domain(x0)))
% 0.20/0.39 = { by axiom 2 (domain2) R->L }
% 0.20/0.39 domain(multiplication(one, x0))
% 0.20/0.39 = { by axiom 1 (multiplicative_left_identity) }
% 0.20/0.39 domain(x0)
% 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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