TSTP Solution File: CSR032+2 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : CSR032+2 : TPTP v8.1.2. Released v3.4.0.
% 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 : n011.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 21:41:07 EDT 2023

% Result   : Theorem 34.16s 4.74s
% Output   : Proof 34.16s
% 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  : CSR032+2 : TPTP v8.1.2. Released v3.4.0.
% 0.07/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34  % Computer : n011.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 : Mon Aug 28 12:53:40 EDT 2023
% 0.12/0.34  % CPUTime  : 
% 34.16/4.74  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 34.16/4.74  
% 34.16/4.74  % SZS status Theorem
% 34.16/4.74  
% 34.16/4.74  % SZS output start Proof
% 34.16/4.74  Take the following subset of the input axioms:
% 34.16/4.74    fof(ax1_1014, axiom, ![X]: (individual(X) => isa(X, c_individual))).
% 34.16/4.74    fof(ax1_153, axiom, ![OBJ]: ~(tptpcol_1_1(OBJ) & tptpcol_1_65536(OBJ))).
% 34.16/4.74    fof(ax1_167, axiom, ![OBJ2]: ~(individual(OBJ2) & setorcollection(OBJ2))).
% 34.16/4.74    fof(ax1_289, axiom, ![OBJ2]: ~(collection(OBJ2) & individual(OBJ2))).
% 34.16/4.74    fof(ax1_3, axiom, ![OBJ2]: ~(intangible(OBJ2) & partiallytangible(OBJ2))).
% 34.16/4.74    fof(ax1_318, axiom, individual(f_citynamedfn(s_agen, c_france))).
% 34.16/4.74    fof(ax1_363, axiom, ![COL1, COL2, OBJ2]: ~(isa(OBJ2, COL1) & (isa(OBJ2, COL2) & disjointwith(COL1, COL2)))).
% 34.16/4.74    fof(ax1_488, axiom, ![OBJ2]: ~(tptpcol_3_98305(OBJ2) & tptpcol_3_114688(OBJ2))).
% 34.16/4.74    fof(ax1_521, axiom, ![X2]: ~affiliatedwith(X2, X2)).
% 34.16/4.74    fof(ax1_698, axiom, ![X2]: ~objectfoundinlocation(X2, X2)).
% 34.16/4.74    fof(ax1_901, axiom, ![X2]: ~borderson(X2, X2)).
% 34.16/4.74    fof(query82, conjecture, ?[COL]: (mtvisible(c_reasoningaboutpossibleantecedentsmt) => isa(f_citynamedfn(s_agen, c_france), COL))).
% 34.16/4.74  
% 34.16/4.74  Now clausify the problem and encode Horn clauses using encoding 3 of
% 34.16/4.74  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 34.16/4.74  We repeatedly replace C & s=t => u=v by the two clauses:
% 34.16/4.74    fresh(y, y, x1...xn) = u
% 34.16/4.74    C => fresh(s, t, x1...xn) = v
% 34.16/4.74  where fresh is a fresh function symbol and x1..xn are the free
% 34.16/4.74  variables of u and v.
% 34.16/4.74  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 34.16/4.74  input problem has no model of domain size 1).
% 34.16/4.74  
% 34.16/4.74  The encoding turns the above axioms into the following unit equations and goals:
% 34.16/4.74  
% 34.16/4.74  Axiom 1 (ax1_318): individual(f_citynamedfn(s_agen, c_france)) = true2.
% 34.16/4.74  Axiom 2 (ax1_1014): fresh805(X, X, Y) = true2.
% 34.16/4.74  Axiom 3 (ax1_1014): fresh805(individual(X), true2, X) = isa(X, c_individual).
% 34.16/4.74  
% 34.16/4.74  Goal 1 (query82_1): isa(f_citynamedfn(s_agen, c_france), X) = true2.
% 34.16/4.74  The goal is true when:
% 34.16/4.74    X = c_individual
% 34.16/4.74  
% 34.16/4.74  Proof:
% 34.16/4.74    isa(f_citynamedfn(s_agen, c_france), c_individual)
% 34.16/4.74  = { by axiom 3 (ax1_1014) R->L }
% 34.16/4.74    fresh805(individual(f_citynamedfn(s_agen, c_france)), true2, f_citynamedfn(s_agen, c_france))
% 34.16/4.74  = { by axiom 1 (ax1_318) }
% 34.16/4.74    fresh805(true2, true2, f_citynamedfn(s_agen, c_france))
% 34.16/4.74  = { by axiom 2 (ax1_1014) }
% 34.16/4.74    true2
% 34.16/4.74  % SZS output end Proof
% 34.16/4.74  
% 34.16/4.74  RESULT: Theorem (the conjecture is true).
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