TSTP Solution File: SET148+3 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SET148+3 : TPTP v8.1.2. Released v2.2.0.
% 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 : n032.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 15:31:28 EDT 2023

% Result   : Theorem 0.10s 0.34s
% Output   : Proof 0.10s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.10  % Problem  : SET148+3 : TPTP v8.1.2. Released v2.2.0.
% 0.00/0.11  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.10/0.30  % Computer : n032.cluster.edu
% 0.10/0.30  % Model    : x86_64 x86_64
% 0.10/0.30  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.10/0.30  % Memory   : 8042.1875MB
% 0.10/0.30  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.10/0.30  % CPULimit : 300
% 0.10/0.30  % WCLimit  : 300
% 0.10/0.30  % DateTime : Sat Aug 26 09:29:43 EDT 2023
% 0.10/0.31  % CPUTime  : 
% 0.10/0.34  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.10/0.34  
% 0.10/0.34  % SZS status Theorem
% 0.10/0.34  
% 0.10/0.34  % SZS output start Proof
% 0.10/0.34  Take the following subset of the input axioms:
% 0.10/0.34    fof(equal_defn, axiom, ![B, C]: (B=C <=> (subset(B, C) & subset(C, B)))).
% 0.10/0.34    fof(prove_idempotency_of_intersection, conjecture, ![B2]: intersection(B2, B2)=B2).
% 0.10/0.34    fof(subset_intersection, axiom, ![B2, C2]: (subset(B2, C2) => intersection(B2, C2)=B2)).
% 0.10/0.34  
% 0.10/0.34  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.10/0.34  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.10/0.34  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.10/0.34    fresh(y, y, x1...xn) = u
% 0.10/0.34    C => fresh(s, t, x1...xn) = v
% 0.10/0.34  where fresh is a fresh function symbol and x1..xn are the free
% 0.10/0.34  variables of u and v.
% 0.10/0.34  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.10/0.34  input problem has no model of domain size 1).
% 0.10/0.34  
% 0.10/0.34  The encoding turns the above axioms into the following unit equations and goals:
% 0.10/0.34  
% 0.10/0.34  Axiom 1 (equal_defn): subset(X, X) = true.
% 0.10/0.34  Axiom 2 (subset_intersection): fresh5(X, X, Y, Z) = Y.
% 0.10/0.34  Axiom 3 (subset_intersection): fresh5(subset(X, Y), true, X, Y) = intersection(X, Y).
% 0.10/0.34  
% 0.10/0.34  Goal 1 (prove_idempotency_of_intersection): intersection(b, b) = b.
% 0.10/0.34  Proof:
% 0.10/0.34    intersection(b, b)
% 0.10/0.34  = { by axiom 3 (subset_intersection) R->L }
% 0.10/0.34    fresh5(subset(b, b), true, b, b)
% 0.10/0.34  = { by axiom 1 (equal_defn) }
% 0.10/0.34    fresh5(true, true, b, b)
% 0.10/0.34  = { by axiom 2 (subset_intersection) }
% 0.10/0.34    b
% 0.10/0.34  % SZS output end Proof
% 0.10/0.34  
% 0.10/0.34  RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------