TSTP Solution File: KLE072+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : KLE072+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 : n025.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 05:35:46 EDT 2023
% Result : Theorem 0.19s 0.40s
% 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
% 0.00/0.12 % Problem : KLE072+1 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.13 % 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 : n025.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 : Tue Aug 29 11:30:10 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.40 Command-line arguments: --no-flatten-goal
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% 0.19/0.40 % SZS status Theorem
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% 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(domain2, axiom, ![X0, X1]: domain(multiplication(X0, X1))=domain(multiplication(X0, domain(X1)))).
% 0.19/0.40 fof(domain5, axiom, ![X0_2, X1_2]: domain(addition(X0_2, X1_2))=addition(domain(X0_2), domain(X1_2))).
% 0.19/0.40 fof(goals, conjecture, ![X2, X0_2, X1_2]: domain(multiplication(addition(X0_2, X1_2), domain(X2)))=addition(domain(multiplication(X0_2, domain(X2))), domain(multiplication(X1_2, domain(X2))))).
% 0.19/0.40 fof(left_distributivity, axiom, ![A, B, C]: multiplication(addition(A, B), C)=addition(multiplication(A, C), multiplication(B, C))).
% 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 (domain2): domain(multiplication(X, Y)) = domain(multiplication(X, domain(Y))).
% 0.19/0.40 Axiom 2 (domain5): domain(addition(X, Y)) = addition(domain(X), domain(Y)).
% 0.19/0.40 Axiom 3 (left_distributivity): multiplication(addition(X, Y), Z) = addition(multiplication(X, Z), multiplication(Y, Z)).
% 0.19/0.40
% 0.19/0.40 Goal 1 (goals): domain(multiplication(addition(x0, x1), domain(x2))) = addition(domain(multiplication(x0, domain(x2))), domain(multiplication(x1, domain(x2)))).
% 0.19/0.40 Proof:
% 0.19/0.40 domain(multiplication(addition(x0, x1), domain(x2)))
% 0.19/0.40 = { by axiom 1 (domain2) R->L }
% 0.19/0.40 domain(multiplication(addition(x0, x1), x2))
% 0.19/0.40 = { by axiom 3 (left_distributivity) }
% 0.19/0.40 domain(addition(multiplication(x0, x2), multiplication(x1, x2)))
% 0.19/0.40 = { by axiom 2 (domain5) }
% 0.19/0.40 addition(domain(multiplication(x0, x2)), domain(multiplication(x1, x2)))
% 0.19/0.40 = { by axiom 1 (domain2) }
% 0.19/0.40 addition(domain(multiplication(x0, x2)), domain(multiplication(x1, domain(x2))))
% 0.19/0.40 = { by axiom 1 (domain2) }
% 0.19/0.40 addition(domain(multiplication(x0, domain(x2))), domain(multiplication(x1, domain(x2))))
% 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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