TSTP Solution File: NUN062+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUN062+2 : TPTP v8.1.2. Released v7.3.0.
% 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 : n007.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 12:51:48 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
% 0.07/0.12 % Problem : NUN062+2 : TPTP v8.1.2. Released v7.3.0.
% 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 : n007.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 : Sun Aug 27 08:51:12 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.20/0.41 Command-line arguments: --ground-connectedness --complete-subsets
% 0.20/0.41
% 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(axiom_1, axiom, ?[Y24]: ![X19]: ((~r1(X19) & X19!=Y24) | (r1(X19) & X19=Y24))).
% 0.20/0.41 fof(axiom_1a, axiom, ![X1, X8]: ?[Y4]: (?[Y5]: (?[Y15]: (r2(X8, Y15) & r3(X1, Y15, Y5)) & Y5=Y4) & ?[Y7]: (r2(Y7, Y4) & r3(X1, X8, Y7)))).
% 0.20/0.41 fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~r1(Y20) | Y20!=Y10) | ~r2(X7, Y10))).
% 0.20/0.41 fof(infiniteNumbers, conjecture, ![X1_2]: ?[Y2, Y1]: (![Y4_2]: (~r1(Y4_2) | Y1!=Y4_2) & ?[Y3]: (r3(X1_2, Y1, Y3) & Y3=Y2))).
% 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:
% 0.20/0.41
% 0.20/0.41 Axiom 1 (infiniteNumbers): fresh(X, X, Y) = Y.
% 0.20/0.41 Axiom 2 (axiom_1_1): fresh12(X, X, Y) = y24.
% 0.20/0.41 Axiom 3 (infiniteNumbers_1): fresh7(X, X, Y) = true2.
% 0.20/0.41 Axiom 4 (axiom_1_1): fresh12(r1(X), true2, X) = X.
% 0.20/0.41 Axiom 5 (axiom_1a_3): r3(X, Y, y7(X, Y)) = true2.
% 0.20/0.41 Axiom 6 (infiniteNumbers): fresh(r3(x1, X, Y), true2, X) = y4(X).
% 0.20/0.41 Axiom 7 (infiniteNumbers_1): fresh7(r3(x1, X, Y), true2, X) = r1(y4(X)).
% 0.20/0.41
% 0.20/0.41 Lemma 8: r1(X) = true2.
% 0.20/0.41 Proof:
% 0.20/0.41 r1(X)
% 0.20/0.41 = { by axiom 1 (infiniteNumbers) R->L }
% 0.20/0.41 r1(fresh(true2, true2, X))
% 0.20/0.41 = { by axiom 5 (axiom_1a_3) R->L }
% 0.20/0.41 r1(fresh(r3(x1, X, y7(x1, X)), true2, X))
% 0.20/0.41 = { by axiom 6 (infiniteNumbers) }
% 0.20/0.41 r1(y4(X))
% 0.20/0.41 = { by axiom 7 (infiniteNumbers_1) R->L }
% 0.20/0.42 fresh7(r3(x1, X, y7(x1, X)), true2, X)
% 0.20/0.42 = { by axiom 5 (axiom_1a_3) }
% 0.20/0.42 fresh7(true2, true2, X)
% 0.20/0.42 = { by axiom 3 (infiniteNumbers_1) }
% 0.20/0.42 true2
% 0.20/0.42
% 0.20/0.42 Goal 1 (true_equals_false): true = false.
% 0.20/0.42 Proof:
% 0.20/0.42 true
% 0.20/0.42 = { by axiom 4 (axiom_1_1) R->L }
% 0.20/0.42 fresh12(r1(true), true2, true)
% 0.20/0.42 = { by lemma 8 }
% 0.20/0.42 fresh12(true2, true2, true)
% 0.20/0.42 = { by axiom 2 (axiom_1_1) }
% 0.20/0.42 y24
% 0.20/0.42 = { by axiom 2 (axiom_1_1) R->L }
% 0.20/0.42 fresh12(true2, true2, false)
% 0.20/0.42 = { by lemma 8 R->L }
% 0.20/0.42 fresh12(r1(false), true2, false)
% 0.20/0.42 = { by axiom 4 (axiom_1_1) }
% 0.20/0.42 false
% 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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