TSTP Solution File: KLE119+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : KLE119+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 : n019.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 05:35:57 EDT 2023
% Result : Theorem 0.13s 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 : KLE119+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 : n019.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 11:57:42 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.13/0.40 Command-line arguments: --no-flatten-goal
% 0.13/0.40
% 0.13/0.40 % SZS status Theorem
% 0.13/0.40
% 0.13/0.40 % SZS output start Proof
% 0.13/0.40 Take the following subset of the input axioms:
% 0.19/0.40 fof(additive_commutativity, axiom, ![A, B]: addition(A, B)=addition(B, A)).
% 0.19/0.40 fof(additive_identity, axiom, ![A2]: addition(A2, zero)=A2).
% 0.19/0.40 fof(backward_diamond, axiom, ![X0, X1]: backward_diamond(X0, X1)=codomain(multiplication(codomain(X1), X0))).
% 0.19/0.40 fof(codomain1, axiom, ![X0_2]: multiplication(X0_2, coantidomain(X0_2))=zero).
% 0.19/0.40 fof(codomain3, axiom, ![X0_2]: addition(coantidomain(coantidomain(X0_2)), coantidomain(X0_2))=one).
% 0.19/0.40 fof(codomain4, axiom, ![X0_2]: codomain(X0_2)=coantidomain(coantidomain(X0_2))).
% 0.19/0.40 fof(goals, conjecture, ![X0_2, X1_2]: addition(backward_diamond(zero, domain(X0_2)), domain(X1_2))=domain(X1_2)).
% 0.19/0.40 fof(multiplicative_left_identity, axiom, ![A2]: multiplication(one, A2)=A2).
% 0.19/0.40 fof(right_annihilation, axiom, ![A2]: multiplication(A2, zero)=zero).
% 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 (codomain4): codomain(X) = coantidomain(coantidomain(X)).
% 0.19/0.40 Axiom 2 (additive_commutativity): addition(X, Y) = addition(Y, X).
% 0.19/0.40 Axiom 3 (additive_identity): addition(X, zero) = X.
% 0.19/0.40 Axiom 4 (right_annihilation): multiplication(X, zero) = zero.
% 0.19/0.40 Axiom 5 (multiplicative_left_identity): multiplication(one, X) = X.
% 0.19/0.40 Axiom 6 (codomain1): multiplication(X, coantidomain(X)) = zero.
% 0.19/0.40 Axiom 7 (backward_diamond): backward_diamond(X, Y) = codomain(multiplication(codomain(Y), X)).
% 0.19/0.40 Axiom 8 (codomain3): addition(coantidomain(coantidomain(X)), coantidomain(X)) = one.
% 0.19/0.40
% 0.19/0.40 Lemma 9: coantidomain(one) = zero.
% 0.19/0.40 Proof:
% 0.19/0.40 coantidomain(one)
% 0.19/0.40 = { by axiom 5 (multiplicative_left_identity) R->L }
% 0.19/0.40 multiplication(one, coantidomain(one))
% 0.19/0.40 = { by axiom 6 (codomain1) }
% 0.19/0.40 zero
% 0.19/0.40
% 0.19/0.40 Goal 1 (goals): addition(backward_diamond(zero, domain(x0)), domain(x1)) = domain(x1).
% 0.19/0.40 Proof:
% 0.19/0.40 addition(backward_diamond(zero, domain(x0)), domain(x1))
% 0.19/0.40 = { by axiom 2 (additive_commutativity) }
% 0.19/0.40 addition(domain(x1), backward_diamond(zero, domain(x0)))
% 0.19/0.40 = { by axiom 7 (backward_diamond) }
% 0.19/0.40 addition(domain(x1), codomain(multiplication(codomain(domain(x0)), zero)))
% 0.19/0.40 = { by axiom 4 (right_annihilation) }
% 0.19/0.40 addition(domain(x1), codomain(zero))
% 0.19/0.40 = { by axiom 2 (additive_commutativity) }
% 0.19/0.40 addition(codomain(zero), domain(x1))
% 0.19/0.40 = { by axiom 1 (codomain4) }
% 0.19/0.40 addition(coantidomain(coantidomain(zero)), domain(x1))
% 0.19/0.40 = { by lemma 9 R->L }
% 0.19/0.40 addition(coantidomain(coantidomain(coantidomain(one))), domain(x1))
% 0.19/0.41 = { by axiom 1 (codomain4) R->L }
% 0.19/0.41 addition(coantidomain(codomain(one)), domain(x1))
% 0.19/0.41 = { by axiom 3 (additive_identity) R->L }
% 0.19/0.41 addition(coantidomain(addition(codomain(one), zero)), domain(x1))
% 0.19/0.41 = { by lemma 9 R->L }
% 0.19/0.41 addition(coantidomain(addition(codomain(one), coantidomain(one))), domain(x1))
% 0.19/0.41 = { by axiom 1 (codomain4) }
% 0.19/0.41 addition(coantidomain(addition(coantidomain(coantidomain(one)), coantidomain(one))), domain(x1))
% 0.19/0.41 = { by axiom 8 (codomain3) }
% 0.19/0.41 addition(coantidomain(one), domain(x1))
% 0.19/0.41 = { by lemma 9 }
% 0.19/0.41 addition(zero, domain(x1))
% 0.19/0.41 = { by axiom 2 (additive_commutativity) R->L }
% 0.19/0.41 addition(domain(x1), zero)
% 0.19/0.41 = { by axiom 3 (additive_identity) }
% 0.19/0.41 domain(x1)
% 0.19/0.41 % SZS output end Proof
% 0.19/0.41
% 0.19/0.41 RESULT: Theorem (the conjecture is true).
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