TSTP Solution File: KLE085+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : KLE085+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 : n029.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 05:35:49 EDT 2023
% Result : Theorem 0.20s 0.40s
% 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
% 0.07/0.12 % Problem : KLE085+1 : TPTP v8.1.2. Released v4.0.0.
% 0.07/0.13 % 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 : n029.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 12:46:41 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.20/0.40 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.20/0.40
% 0.20/0.40 % SZS status Theorem
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% 0.20/0.40 % SZS output start Proof
% 0.20/0.40 Take the following subset of the input axioms:
% 0.20/0.40 fof(additive_associativity, axiom, ![A, B, C]: addition(A, addition(B, C))=addition(addition(A, B), C)).
% 0.20/0.40 fof(additive_idempotence, axiom, ![A2]: addition(A2, A2)=A2).
% 0.20/0.40 fof(domain3, axiom, ![X0]: addition(antidomain(antidomain(X0)), antidomain(X0))=one).
% 0.20/0.40 fof(domain4, axiom, ![X0_2]: domain(X0_2)=antidomain(antidomain(X0_2))).
% 0.20/0.40 fof(goals, conjecture, ![X0_2]: addition(domain(X0_2), one)=one).
% 0.20/0.40
% 0.20/0.40 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.40 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.40 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.40 fresh(y, y, x1...xn) = u
% 0.20/0.40 C => fresh(s, t, x1...xn) = v
% 0.20/0.40 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.40 variables of u and v.
% 0.20/0.40 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.40 input problem has no model of domain size 1).
% 0.20/0.40
% 0.20/0.40 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.40
% 0.20/0.40 Axiom 1 (domain4): domain(X) = antidomain(antidomain(X)).
% 0.20/0.40 Axiom 2 (additive_idempotence): addition(X, X) = X.
% 0.20/0.40 Axiom 3 (additive_associativity): addition(X, addition(Y, Z)) = addition(addition(X, Y), Z).
% 0.20/0.40 Axiom 4 (domain3): addition(antidomain(antidomain(X)), antidomain(X)) = one.
% 0.20/0.40
% 0.20/0.40 Lemma 5: addition(domain(X), antidomain(X)) = one.
% 0.20/0.40 Proof:
% 0.20/0.40 addition(domain(X), antidomain(X))
% 0.20/0.40 = { by axiom 1 (domain4) }
% 0.20/0.40 addition(antidomain(antidomain(X)), antidomain(X))
% 0.20/0.40 = { by axiom 4 (domain3) }
% 0.20/0.40 one
% 0.20/0.40
% 0.20/0.40 Goal 1 (goals): addition(domain(x0), one) = one.
% 0.20/0.40 Proof:
% 0.20/0.40 addition(domain(x0), one)
% 0.20/0.40 = { by lemma 5 R->L }
% 0.20/0.40 addition(domain(x0), addition(domain(x0), antidomain(x0)))
% 0.20/0.40 = { by axiom 3 (additive_associativity) }
% 0.20/0.40 addition(addition(domain(x0), domain(x0)), antidomain(x0))
% 0.20/0.40 = { by axiom 2 (additive_idempotence) }
% 0.20/0.40 addition(domain(x0), antidomain(x0))
% 0.20/0.40 = { by lemma 5 }
% 0.20/0.40 one
% 0.20/0.40 % SZS output end Proof
% 0.20/0.40
% 0.20/0.40 RESULT: Theorem (the conjecture is true).
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