TSTP Solution File: SWB002+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SWB002+2 : TPTP v8.1.2. Released v5.2.0.
% 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 20:12:40 EDT 2023
% Result : Theorem 0.17s 0.39s
% Output : Proof 0.17s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.10/0.14 % Problem : SWB002+2 : TPTP v8.1.2. Released v5.2.0.
% 0.10/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.11/0.35 % Computer : n025.cluster.edu
% 0.11/0.35 % Model : x86_64 x86_64
% 0.11/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.35 % Memory : 8042.1875MB
% 0.11/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.11/0.35 % CPULimit : 300
% 0.11/0.35 % WCLimit : 300
% 0.11/0.35 % DateTime : Sun Aug 27 07:35:38 EDT 2023
% 0.11/0.35 % CPUTime :
% 0.17/0.39 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.17/0.39
% 0.17/0.39 % SZS status Theorem
% 0.17/0.39
% 0.17/0.39 % SZS output start Proof
% 0.17/0.39 Take the following subset of the input axioms:
% 0.17/0.39 fof(testcase_conclusion_fullish_002_Existential_Blank_Nodes, conjecture, ?[BNODE_x, BNODE_y]: (iext(uri_ex_p, BNODE_x, BNODE_y) & iext(uri_ex_q, BNODE_y, BNODE_x))).
% 0.17/0.39 fof(testcase_premise_fullish_002_Existential_Blank_Nodes, axiom, ?[BNODE_o]: (iext(uri_ex_p, uri_ex_s, BNODE_o) & iext(uri_ex_q, BNODE_o, uri_ex_s))).
% 0.17/0.39
% 0.17/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.17/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.17/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.17/0.39 fresh(y, y, x1...xn) = u
% 0.17/0.39 C => fresh(s, t, x1...xn) = v
% 0.17/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.17/0.39 variables of u and v.
% 0.17/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.17/0.39 input problem has no model of domain size 1).
% 0.17/0.39
% 0.17/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.17/0.39
% 0.17/0.39 Axiom 1 (testcase_premise_fullish_002_Existential_Blank_Nodes): iext(uri_ex_p, uri_ex_s, bnode_o) = true2.
% 0.17/0.39 Axiom 2 (testcase_premise_fullish_002_Existential_Blank_Nodes_1): iext(uri_ex_q, bnode_o, uri_ex_s) = true2.
% 0.17/0.39
% 0.17/0.39 Goal 1 (testcase_conclusion_fullish_002_Existential_Blank_Nodes): tuple(iext(uri_ex_p, X, Y), iext(uri_ex_q, Y, X)) = tuple(true2, true2).
% 0.17/0.39 The goal is true when:
% 0.17/0.39 X = uri_ex_s
% 0.17/0.39 Y = bnode_o
% 0.17/0.39
% 0.17/0.39 Proof:
% 0.17/0.39 tuple(iext(uri_ex_p, uri_ex_s, bnode_o), iext(uri_ex_q, bnode_o, uri_ex_s))
% 0.17/0.39 = { by axiom 1 (testcase_premise_fullish_002_Existential_Blank_Nodes) }
% 0.17/0.39 tuple(true2, iext(uri_ex_q, bnode_o, uri_ex_s))
% 0.17/0.39 = { by axiom 2 (testcase_premise_fullish_002_Existential_Blank_Nodes_1) }
% 0.17/0.39 tuple(true2, true2)
% 0.17/0.39 % SZS output end Proof
% 0.17/0.39
% 0.17/0.39 RESULT: Theorem (the conjecture is true).
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