TSTP Solution File: KLE006+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : KLE006+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 : n011.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 05:35:23 EDT 2023
% Result : Theorem 0.20s 0.41s
% 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
% 0.00/0.12 % Problem : KLE006+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
% 0.14/0.35 % Computer : n011.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 11:33:26 EDT 2023
% 0.14/0.35 % CPUTime :
% 0.20/0.41 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
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% 0.20/0.41 % SZS status Theorem
% 0.20/0.41
% 0.20/0.41 % SZS output start Proof
% 0.20/0.41 Take the following subset of the input axioms:
% 0.20/0.41 fof(additive_commutativity, axiom, ![A, B]: addition(A, B)=addition(B, A)).
% 0.20/0.41 fof(goals, conjecture, ![X0]: (test(X0) => one=addition(X0, c(X0)))).
% 0.20/0.41 fof(test_2, axiom, ![X1, X0_2]: (complement(X1, X0_2) <=> (multiplication(X0_2, X1)=zero & (multiplication(X1, X0_2)=zero & addition(X0_2, X1)=one)))).
% 0.20/0.41 fof(test_3, axiom, ![X0_2, X1_2]: (test(X0_2) => (c(X0_2)=X1_2 <=> complement(X0_2, X1_2)))).
% 0.20/0.41
% 0.20/0.41 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.41 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.41 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.41 fresh(y, y, x1...xn) = u
% 0.20/0.41 C => fresh(s, t, x1...xn) = v
% 0.20/0.41 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.41 variables of u and v.
% 0.20/0.41 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.41 input problem has no model of domain size 1).
% 0.20/0.41
% 0.20/0.41 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.42
% 0.20/0.42 Axiom 1 (goals): test(x0) = true.
% 0.20/0.42 Axiom 2 (additive_commutativity): addition(X, Y) = addition(Y, X).
% 0.20/0.42 Axiom 3 (test_2_1): fresh8(X, X, Y, Z) = one.
% 0.20/0.42 Axiom 4 (test_3): fresh5(X, X, Y, Z) = complement(Y, Z).
% 0.20/0.42 Axiom 5 (test_3): fresh4(X, X, Y, Z) = true.
% 0.20/0.42 Axiom 6 (test_3): fresh5(test(X), true, X, Y) = fresh4(c(X), Y, X, Y).
% 0.20/0.42 Axiom 7 (test_2_1): fresh8(complement(X, Y), true, Y, X) = addition(Y, X).
% 0.20/0.42
% 0.20/0.42 Goal 1 (goals_1): one = addition(x0, c(x0)).
% 0.20/0.42 Proof:
% 0.20/0.42 one
% 0.20/0.42 = { by axiom 3 (test_2_1) R->L }
% 0.20/0.42 fresh8(true, true, c(x0), x0)
% 0.20/0.42 = { by axiom 5 (test_3) R->L }
% 0.20/0.42 fresh8(fresh4(c(x0), c(x0), x0, c(x0)), true, c(x0), x0)
% 0.20/0.42 = { by axiom 6 (test_3) R->L }
% 0.20/0.42 fresh8(fresh5(test(x0), true, x0, c(x0)), true, c(x0), x0)
% 0.20/0.42 = { by axiom 1 (goals) }
% 0.20/0.42 fresh8(fresh5(true, true, x0, c(x0)), true, c(x0), x0)
% 0.20/0.42 = { by axiom 4 (test_3) }
% 0.20/0.42 fresh8(complement(x0, c(x0)), true, c(x0), x0)
% 0.20/0.42 = { by axiom 7 (test_2_1) }
% 0.20/0.42 addition(c(x0), x0)
% 0.20/0.42 = { by axiom 2 (additive_commutativity) }
% 0.20/0.42 addition(x0, c(x0))
% 0.20/0.42 % SZS output end Proof
% 0.20/0.42
% 0.20/0.42 RESULT: Theorem (the conjecture is true).
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