TSTP Solution File: AGT017+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : AGT017+2 : TPTP v8.1.2. Bugfixed 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 : n026.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 : Wed Aug 30 15:55:44 EDT 2023
% Result : Theorem 8.10s 1.40s
% Output : Proof 8.10s
% 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 : AGT017+2 : TPTP v8.1.2. Bugfixed 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
% 0.13/0.34 % Computer : n026.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 17:22:49 EDT 2023
% 0.13/0.34 % CPUTime :
% 8.10/1.40 Command-line arguments: --no-flatten-goal
% 8.10/1.40
% 8.10/1.40 % SZS status Theorem
% 8.10/1.40
% 8.10/1.49 % SZS output start Proof
% 8.10/1.49 Take the following subset of the input axioms:
% 8.10/1.49 fof(deduced_366, axiom, ~accept_city(muslimcountrybhumanitarianorganization, towna)).
% 8.10/1.49 fof(query_17, conjecture, ?[X]: ~accept_city(muslimcountrybhumanitarianorganization, X)).
% 8.10/1.49
% 8.10/1.49 Now clausify the problem and encode Horn clauses using encoding 3 of
% 8.10/1.49 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 8.10/1.49 We repeatedly replace C & s=t => u=v by the two clauses:
% 8.10/1.49 fresh(y, y, x1...xn) = u
% 8.10/1.49 C => fresh(s, t, x1...xn) = v
% 8.10/1.49 where fresh is a fresh function symbol and x1..xn are the free
% 8.10/1.49 variables of u and v.
% 8.10/1.49 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 8.10/1.49 input problem has no model of domain size 1).
% 8.10/1.49
% 8.10/1.49 The encoding turns the above axioms into the following unit equations and goals:
% 8.10/1.49
% 8.10/1.49 Axiom 1 (query_17): accept_city(muslimcountrybhumanitarianorganization, X) = true2.
% 8.10/1.49
% 8.10/1.49 Goal 1 (deduced_366): accept_city(muslimcountrybhumanitarianorganization, towna) = true2.
% 8.10/1.49 Proof:
% 8.10/1.49 accept_city(muslimcountrybhumanitarianorganization, towna)
% 8.10/1.49 = { by axiom 1 (query_17) }
% 8.10/1.49 true2
% 8.10/1.49 % SZS output end Proof
% 8.10/1.49
% 8.10/1.49 RESULT: Theorem (the conjecture is true).
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