TSTP Solution File: NUM322+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM322+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 : n009.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 11:55:54 EDT 2023
% Result : Theorem 9.86s 1.63s
% Output : Proof 9.86s
% 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 : NUM322+1 : TPTP v8.1.2. Released v3.1.0.
% 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 : n009.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 : Fri Aug 25 10:56:21 EDT 2023
% 0.20/0.34 % CPUTime :
% 9.86/1.63 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 9.86/1.63
% 9.86/1.63 % SZS status Theorem
% 9.86/1.63
% 9.86/1.64 % SZS output start Proof
% 9.86/1.64 Take the following subset of the input axioms:
% 9.86/1.64 fof(less_property, axiom, ![X, Y]: (less(X, Y) <=> (~less(Y, X) & Y!=X))).
% 9.86/1.64 fof(rdn5, axiom, rdn_translate(n5, rdn_pos(rdnn(n5)))).
% 9.86/1.64 fof(rdnn5, axiom, rdn_translate(nn5, rdn_neg(rdnn(n5)))).
% 9.86/1.64 fof(sum_entry_point_posx_negx, axiom, ![RDN_X, POS_X, NEG_X]: ((rdn_translate(POS_X, rdn_pos(RDN_X)) & rdn_translate(NEG_X, rdn_neg(RDN_X))) => sum(POS_X, NEG_X, n0))).
% 9.86/1.64 fof(sum_n5_nn5_what, conjecture, ?[X2]: sum(n5, nn5, X2)).
% 9.86/1.64
% 9.86/1.64 Now clausify the problem and encode Horn clauses using encoding 3 of
% 9.86/1.64 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 9.86/1.64 We repeatedly replace C & s=t => u=v by the two clauses:
% 9.86/1.64 fresh(y, y, x1...xn) = u
% 9.86/1.64 C => fresh(s, t, x1...xn) = v
% 9.86/1.64 where fresh is a fresh function symbol and x1..xn are the free
% 9.86/1.64 variables of u and v.
% 9.86/1.64 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 9.86/1.64 input problem has no model of domain size 1).
% 9.86/1.64
% 9.86/1.64 The encoding turns the above axioms into the following unit equations and goals:
% 9.86/1.64
% 9.86/1.64 Axiom 1 (rdn5): rdn_translate(n5, rdn_pos(rdnn(n5))) = true2.
% 9.86/1.64 Axiom 2 (rdnn5): rdn_translate(nn5, rdn_neg(rdnn(n5))) = true2.
% 9.86/1.64 Axiom 3 (sum_entry_point_posx_negx): fresh7(X, X, Y, Z) = true2.
% 9.86/1.64 Axiom 4 (sum_entry_point_posx_negx): fresh8(X, X, Y, Z, W) = sum(Y, Z, n0).
% 9.86/1.64 Axiom 5 (sum_entry_point_posx_negx): fresh8(rdn_translate(X, rdn_neg(Y)), true2, Z, X, Y) = fresh7(rdn_translate(Z, rdn_pos(Y)), true2, Z, X).
% 9.86/1.64
% 9.86/1.64 Goal 1 (sum_n5_nn5_what): sum(n5, nn5, X) = true2.
% 9.86/1.64 The goal is true when:
% 9.86/1.64 X = n0
% 9.86/1.64
% 9.86/1.64 Proof:
% 9.86/1.64 sum(n5, nn5, n0)
% 9.86/1.64 = { by axiom 4 (sum_entry_point_posx_negx) R->L }
% 9.86/1.64 fresh8(true2, true2, n5, nn5, rdnn(n5))
% 9.86/1.64 = { by axiom 2 (rdnn5) R->L }
% 9.86/1.64 fresh8(rdn_translate(nn5, rdn_neg(rdnn(n5))), true2, n5, nn5, rdnn(n5))
% 9.86/1.64 = { by axiom 5 (sum_entry_point_posx_negx) }
% 9.86/1.64 fresh7(rdn_translate(n5, rdn_pos(rdnn(n5))), true2, n5, nn5)
% 9.86/1.64 = { by axiom 1 (rdn5) }
% 9.86/1.64 fresh7(true2, true2, n5, nn5)
% 9.86/1.64 = { by axiom 3 (sum_entry_point_posx_negx) }
% 9.86/1.64 true2
% 9.86/1.64 % SZS output end Proof
% 9.86/1.64
% 9.86/1.64 RESULT: Theorem (the conjecture is true).
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