TPTP Problem File: NUM927_1.p
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- Solve Problem
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% File : NUM927_1 : TPTP v8.2.0. Bugfixed v6.4.0.
% Domain : Number Theory
% Problem : The Collatz Conjecture
% Version : Especial.
% English : f(X) = 3X + 1 if X is odd, X/2 if X is even. Prove this is cyclic.
% e.g., 3,10,5,16,8,4,2,1,4,2,1
% Refs :
% Source : [TPTP]
% Names :
% Status : Open
% Rating : 1.00 v6.4.0
% Syntax : Number of formulae : 7 ( 0 unt; 2 typ; 0 def)
% Number of atoms : 10 ( 8 equ)
% Maximal formula atoms : 2 ( 1 avg)
% Number of connectives : 5 ( 0 ~; 0 |; 0 &)
% ( 0 <=>; 5 =>; 0 <=; 0 <~>)
% Maximal formula depth : 4 ( 4 avg)
% Maximal term depth : 3 ( 1 avg)
% Number arithmetic : 28 ( 2 atm; 6 fun; 12 num; 8 var)
% Number of types : 1 ( 0 usr; 1 ari)
% Number of type conns : 3 ( 2 >; 1 *; 0 +; 0 <<)
% Number of predicates : 3 ( 0 usr; 0 prp; 2-2 aty)
% Number of functors : 11 ( 2 usr; 4 con; 0-2 aty)
% Number of variables : 8 ( 7 !; 1 ?; 8 :)
% SPC : TF0_OPN_EQU_ARI
% Comments :
% Bugfixes : v5.5.1 - Added types for f and iterate_f.
% : v6.4.0 - Removed alternative conjecture
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tff(f_type,type,
f: $int > $int ).
tff(iterate_f_type,type,
iterate_f: ( $int * $int ) > $int ).
tff(f_odd,axiom,
! [X: $int] :
( ( $remainder_t(X,2) = 1 )
=> ( f(X) = $sum($product(3,X),1) ) ) ).
tff(f_even,axiom,
! [X: $int] :
( ( $remainder_t(X,2) = 0 )
=> ( f(X) = $quotient_t(X,2) ) ) ).
tff(iterate_f_base,axiom,
! [I: $int,X: $int] :
( ( I = 1 )
=> ( iterate_f(I,X) = f(X) ) ) ).
tff(iterate_f,axiom,
! [I: $int,X: $int] :
( $greater(I,1)
=> ( iterate_f(I,X) = iterate_f($difference(I,1),f(X)) ) ) ).
tff(iterates_to_1,conjecture,
! [X: $int] :
( $greatereq(X,1)
=> ? [I: $int] : iterate_f(I,X) = 1 ) ).
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