TSTP Solution File: SWB003+3 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWB003+3 : 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 : n013.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 20:12:41 EDT 2023
% Result : Theorem 6.18s 1.18s
% Output : Proof 6.36s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : SWB003+3 : TPTP v8.1.2. Released v5.2.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35 % Computer : n013.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Sun Aug 27 06:25:02 EDT 2023
% 0.13/0.35 % CPUTime :
% 6.18/1.18 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 6.18/1.18
% 6.18/1.18 % SZS status Theorem
% 6.18/1.18
% 6.18/1.18 % SZS output start Proof
% 6.18/1.18 Take the following subset of the input axioms:
% 6.18/1.18 fof(owl_bool_complementof_class, axiom, ![Z, C]: (iext(uri_owl_complementOf, Z, C) => (ic(Z) & (ic(C) & ![X]: (icext(Z, X) <=> ~icext(C, X)))))).
% 6.18/1.18 fof(owl_bool_unionof_class_000, axiom, ![Z2]: (iext(uri_owl_unionOf, Z2, uri_rdf_nil) <=> (ic(Z2) & ![X2]: ~icext(Z2, X2)))).
% 6.18/1.18 fof(owl_class_nothing_ext, axiom, ![X2]: ~icext(uri_owl_Nothing, X2)).
% 6.18/1.18 fof(testcase_conclusion_fullish_003_Blank_Nodes_for_Literals, conjecture, ?[BNODE_x]: iext(uri_ex_p, uri_ex_s, BNODE_x)).
% 6.18/1.18 fof(testcase_premise_fullish_003_Blank_Nodes_for_Literals, axiom, iext(uri_ex_p, uri_ex_s, literal_plain(dat_str_foo))).
% 6.18/1.18
% 6.18/1.18 Now clausify the problem and encode Horn clauses using encoding 3 of
% 6.18/1.18 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 6.36/1.18 We repeatedly replace C & s=t => u=v by the two clauses:
% 6.36/1.18 fresh(y, y, x1...xn) = u
% 6.36/1.18 C => fresh(s, t, x1...xn) = v
% 6.36/1.18 where fresh is a fresh function symbol and x1..xn are the free
% 6.36/1.18 variables of u and v.
% 6.36/1.18 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 6.36/1.18 input problem has no model of domain size 1).
% 6.36/1.18
% 6.36/1.18 The encoding turns the above axioms into the following unit equations and goals:
% 6.36/1.18
% 6.36/1.18 Axiom 1 (testcase_premise_fullish_003_Blank_Nodes_for_Literals): iext(uri_ex_p, uri_ex_s, literal_plain(dat_str_foo)) = true2.
% 6.36/1.18
% 6.36/1.18 Goal 1 (testcase_conclusion_fullish_003_Blank_Nodes_for_Literals): iext(uri_ex_p, uri_ex_s, X) = true2.
% 6.36/1.18 The goal is true when:
% 6.36/1.18 X = literal_plain(dat_str_foo)
% 6.36/1.18
% 6.36/1.18 Proof:
% 6.36/1.18 iext(uri_ex_p, uri_ex_s, literal_plain(dat_str_foo))
% 6.36/1.18 = { by axiom 1 (testcase_premise_fullish_003_Blank_Nodes_for_Literals) }
% 6.36/1.18 true2
% 6.36/1.18 % SZS output end Proof
% 6.36/1.18
% 6.36/1.18 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------