TSTP Solution File: KLE074+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : KLE074+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 : n021.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 05:35:47 EDT 2023
% Result : Theorem 0.20s 0.39s
% 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
% 0.10/0.12 % Problem : KLE074+1 : TPTP v8.1.2. Released v4.0.0.
% 0.10/0.13 % 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 : n021.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 11:22:59 EDT 2023
% 0.14/0.34 % CPUTime :
% 0.20/0.39 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.20/0.39
% 0.20/0.39 % SZS status Theorem
% 0.20/0.39
% 0.20/0.39 % SZS output start Proof
% 0.20/0.39 Take the following subset of the input axioms:
% 0.20/0.39 fof(domain2, axiom, ![X0, X1]: domain(multiplication(X0, X1))=domain(multiplication(X0, domain(X1)))).
% 0.20/0.39 fof(goals, conjecture, ![X2, X0_2, X1_2]: domain(multiplication(multiplication(X0_2, X1_2), domain(X2)))=domain(multiplication(X0_2, domain(multiplication(X1_2, domain(X2)))))).
% 0.20/0.39 fof(multiplicative_associativity, axiom, ![A, B, C]: multiplication(A, multiplication(B, C))=multiplication(multiplication(A, B), C)).
% 0.20/0.39
% 0.20/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.39 fresh(y, y, x1...xn) = u
% 0.20/0.39 C => fresh(s, t, x1...xn) = v
% 0.20/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.39 variables of u and v.
% 0.20/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.39 input problem has no model of domain size 1).
% 0.20/0.39
% 0.20/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.39
% 0.20/0.39 Axiom 1 (domain2): domain(multiplication(X, Y)) = domain(multiplication(X, domain(Y))).
% 0.20/0.39 Axiom 2 (multiplicative_associativity): multiplication(X, multiplication(Y, Z)) = multiplication(multiplication(X, Y), Z).
% 0.20/0.39
% 0.20/0.39 Goal 1 (goals): domain(multiplication(multiplication(x0, x1), domain(x2))) = domain(multiplication(x0, domain(multiplication(x1, domain(x2))))).
% 0.20/0.39 Proof:
% 0.20/0.39 domain(multiplication(multiplication(x0, x1), domain(x2)))
% 0.20/0.39 = { by axiom 1 (domain2) R->L }
% 0.20/0.39 domain(multiplication(multiplication(x0, x1), x2))
% 0.20/0.39 = { by axiom 2 (multiplicative_associativity) R->L }
% 0.20/0.39 domain(multiplication(x0, multiplication(x1, x2)))
% 0.20/0.39 = { by axiom 1 (domain2) }
% 0.20/0.39 domain(multiplication(x0, domain(multiplication(x1, x2))))
% 0.20/0.39 = { by axiom 1 (domain2) }
% 0.20/0.39 domain(multiplication(x0, domain(multiplication(x1, domain(x2)))))
% 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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