set IT = elementary_tree n;
elementary_tree n, Seg (n + 1) are_equipotent
proof
defpred S1[
object ,
object ]
means ( (
n = {} &
c2 = 1 ) or ex
n being
Nat st
(
n = <*n*> &
c2 = n + 2 ) );
A1:
for
x,
y,
z being
set st
x in elementary_tree n &
S1[
x,
y] &
S1[
x,
z] holds
y = z
A5:
for
x being
object st
x in elementary_tree n holds
ex
y being
object st
S1[
x,
y]
consider f being
Function such that A10:
(
dom f = elementary_tree n & ( for
x being
object st
x in elementary_tree n holds
S1[
x,
f . x] ) )
from CLASSES1:sch 1(A5);
take
f
;
WELLORD2:def 4 ( f is one-to-one & dom f = elementary_tree n & rng f = Seg (n + 1) )
thus
f is
one-to-one
( dom f = elementary_tree n & rng f = Seg (n + 1) )proof
let x,
y be
object ;
FUNCT_1:def 4 ( not x in dom f or not y in dom f or not f . x = f . y or x = y )
assume that A11:
x in dom f
and A12:
y in dom f
and A13:
f . x = f . y
;
x = y
A14:
( (
x = {} &
f . x = 1 ) or ex
n being
Nat st
(
x = <*n*> &
f . x = n + 2 ) )
by A10, A11;
A15:
( (
y = {} &
f . y = 1 ) or ex
n being
Nat st
(
y = <*n*> &
f . y = n + 2 ) )
by A10, A12;
A16:
now ( x = {} & f . x = 1 implies for n being Nat holds
( not y = <*n*> or not f . y = n + 2 ) )assume that
x = {}
and A17:
f . x = 1
;
for n being Nat holds
( not y = <*n*> or not f . y = n + 2 )given n being
Nat such that
y = <*n*>
and A18:
f . y = n + 2
;
contradiction
0 + 1
= (n + 1) + 1
by A13, A17, A18;
hence
contradiction
;
verum end;
now ( y = {} & f . y = 1 implies for n being Nat holds
( not x = <*n*> or not f . x = n + 2 ) )assume that
y = {}
and A19:
f . y = 1
;
for n being Nat holds
( not x = <*n*> or not f . x = n + 2 )given n being
Nat such that
x = <*n*>
and A20:
f . x = n + 2
;
contradiction
0 + 1
= (n + 1) + 1
by A13, A19, A20;
hence
contradiction
;
verum end;
hence
x = y
by A13, A14, A15, A16;
verum
end;
thus
dom f = elementary_tree n
by A10;
rng f = Seg (n + 1)
thus
rng f c= Seg (n + 1)
XBOOLE_0:def 10 Seg (n + 1) is_a_prefix_of rng f
let x be
object ;
TARSKI:def 3 ( not x in Seg (n + 1) or x in rng f )
assume A27:
x in Seg (n + 1)
;
x in rng f
then reconsider k =
x as
Nat ;
A28:
1
<= k
by A27, FINSEQ_1:1;
A29:
k <= n + 1
by A27, FINSEQ_1:1;
{} in {{}}
by TARSKI:def 1;
then A30:
{} in elementary_tree n
by XBOOLE_0:def 3;
then
S1[
{} ,
f . {}]
by A10;
then A31:
1
in rng f
by A10, A30, FUNCT_1:def 3;
hence
x in rng f
by A28, A31, XXREAL_0:1;
verum
end;
hence
elementary_tree n is finite
by CARD_1:38; verum