TSTP Solution File: NUM300+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM300+1 : TPTP v8.1.2. Released v3.1.0.
% 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 11:55:51 EDT 2023
% Result : Theorem 7.23s 1.39s
% Output : Proof 7.23s
% 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 : NUM300+1 : TPTP v8.1.2. Released v3.1.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34 % Computer : n025.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Fri Aug 25 15:39:52 EDT 2023
% 0.12/0.34 % CPUTime :
% 7.23/1.39 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 7.23/1.39
% 7.23/1.39 % SZS status Theorem
% 7.23/1.39
% 7.23/1.39 % SZS output start Proof
% 7.23/1.39 Take the following subset of the input axioms:
% 7.23/1.39 fof(less_entry_point_neg_pos, axiom, ![X, Y, RDN_X, RDN_Y]: ((rdn_translate(X, rdn_neg(RDN_X)) & rdn_translate(Y, rdn_pos(RDN_Y))) => less(X, Y))).
% 7.23/1.39 fof(less_property, axiom, ![X2, Y2]: (less(X2, Y2) <=> (~less(Y2, X2) & Y2!=X2))).
% 7.23/1.39 fof(rdn0, axiom, rdn_translate(n0, rdn_pos(rdnn(n0)))).
% 7.23/1.39 fof(rdnn1, axiom, rdn_translate(nn1, rdn_neg(rdnn(n1)))).
% 7.23/1.39 fof(something_less_n0, conjecture, ?[X2]: less(X2, n0)).
% 7.23/1.39
% 7.23/1.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 7.23/1.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 7.23/1.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 7.23/1.39 fresh(y, y, x1...xn) = u
% 7.23/1.39 C => fresh(s, t, x1...xn) = v
% 7.23/1.39 where fresh is a fresh function symbol and x1..xn are the free
% 7.23/1.39 variables of u and v.
% 7.23/1.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 7.23/1.39 input problem has no model of domain size 1).
% 7.23/1.39
% 7.23/1.39 The encoding turns the above axioms into the following unit equations and goals:
% 7.23/1.39
% 7.23/1.39 Axiom 1 (rdn0): rdn_translate(n0, rdn_pos(rdnn(n0))) = true2.
% 7.23/1.39 Axiom 2 (rdnn1): rdn_translate(nn1, rdn_neg(rdnn(n1))) = true2.
% 7.23/1.39 Axiom 3 (less_entry_point_neg_pos): fresh24(X, X, Y, Z) = true2.
% 7.23/1.39 Axiom 4 (less_entry_point_neg_pos): fresh25(X, X, Y, Z, W) = less(Y, Z).
% 7.23/1.39 Axiom 5 (less_entry_point_neg_pos): fresh25(rdn_translate(X, rdn_pos(Y)), true2, Z, X, W) = fresh24(rdn_translate(Z, rdn_neg(W)), true2, Z, X).
% 7.23/1.39
% 7.23/1.39 Goal 1 (something_less_n0): less(X, n0) = true2.
% 7.23/1.39 The goal is true when:
% 7.23/1.39 X = nn1
% 7.23/1.39
% 7.23/1.39 Proof:
% 7.23/1.39 less(nn1, n0)
% 7.23/1.39 = { by axiom 4 (less_entry_point_neg_pos) R->L }
% 7.23/1.39 fresh25(true2, true2, nn1, n0, rdnn(n1))
% 7.23/1.39 = { by axiom 1 (rdn0) R->L }
% 7.23/1.39 fresh25(rdn_translate(n0, rdn_pos(rdnn(n0))), true2, nn1, n0, rdnn(n1))
% 7.23/1.39 = { by axiom 5 (less_entry_point_neg_pos) }
% 7.23/1.39 fresh24(rdn_translate(nn1, rdn_neg(rdnn(n1))), true2, nn1, n0)
% 7.23/1.39 = { by axiom 2 (rdnn1) }
% 7.23/1.39 fresh24(true2, true2, nn1, n0)
% 7.23/1.39 = { by axiom 3 (less_entry_point_neg_pos) }
% 7.23/1.39 true2
% 7.23/1.39 % SZS output end Proof
% 7.23/1.39
% 7.23/1.39 RESULT: Theorem (the conjecture is true).
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