let n be Element of NAT ; ( dom (dyad n) = Seg ((2 |^ n) + 1) & rng (dyad n) = dyadic n )
A1:
dom (dyad n) = Seg ((2 |^ n) + 1)
by Def6;
for x being set holds
( x in rng (dyad n) iff x in dyadic n )
proof
let x be
set ;
( x in rng (dyad n) iff x in dyadic n )
A2:
(
x in rng (dyad n) implies
x in dyadic n )
proof
assume
x in rng (dyad n)
;
x in dyadic n
then consider y being
set such that A3:
y in dom (dyad n)
and A4:
x = (dyad n) . y
by FUNCT_1:def 5;
A5:
y in Seg ((2 |^ n) + 1)
by A3, Def6;
reconsider y =
y as
Element of
NAT by A3;
A6:
1
<= y
by A5, FINSEQ_1:3;
then A7:
1
- 1
<= y - 1
by XREAL_1:15;
y <= (2 |^ n) + 1
by A5, FINSEQ_1:3;
then A8:
y - 1
<= ((2 |^ n) + 1) - 1
by XREAL_1:15;
consider i being
Nat such that A9:
1
+ i = y
by A6, NAT_1:10;
(
i in NAT &
x = (y - 1) / (2 |^ n) )
by A3, A4, Def6, ORDINAL1:def 13;
hence
x in dyadic n
by A7, A8, A9, Def3;
verum
end;
(
x in dyadic n implies
x in rng (dyad n) )
hence
(
x in rng (dyad n) iff
x in dyadic n )
by A2;
verum
end;
hence
( dom (dyad n) = Seg ((2 |^ n) + 1) & rng (dyad n) = dyadic n )
by Def6, TARSKI:2; verum