2
(
∃
x)(
∀
z)(z
∈
x
↔
(z
≤
y
∧
(z = y
1
∨
...
∨
z = y
k
))).
The last claim is obvious. QED
DEFINITION 5.7.5. We write x
≈
<y,z> if and only if there
exists a,b such that
i) x
≈
{a,b};
ii) a
≈
{y};
iii) b
≈
{y,z}.
LEMMA 5.7.2. If x
≈
<y,z>
∧
w
∈
x, then w
≈
{y}
∨
w
≈
{y,z}.
If x
≈
<y,z>
∧
x
≈
<u,v>, then y = u
∧
z = v. For all y,z,
there exists x
≈
<y,z>.
Proof: For the first claim, let x,y,z,w be as given. Let
a,b be such that x
≈
{a,b}, a
≈
{y}, b
≈
{y,z}. Then w = a
∨
w = b. Hence w
≈
{y}
∨
w
≈
{y,z}.
For the second claim, let x
≈
<y,z>, x
≈
<u,v>. Let
x
≈
{a,b}, a
≈
{y}, b
≈
{y,z}
x
≈
{c,d}, c
≈
{u}, d
≈
{u,v}.
Then
(a = c
∨
a = d)
∧
(b = c
∨
b = d)
∧
(c = a
∨
c = b)
∧
(d =
a
∨
d = b).
Since a = c
∨
a = d, we have y = u
∨
(y = u = v). Hence y =
u.
We have b
≈
{y,z}, d
≈
{y,v}. If b = d then z = v. So we can
assume b
≠
d. Hence b = c, d = a. Therefore u = y = z, y = u
= v.
For the third claim, let y,z. By Lemma 5.7.1, let a
≈
{y}
and b
≈
{y,z}. Let x
≈
{a,b}. Then x
≈
<y,z>. QED
DEFINITION 5.7.6. Let k
≥
2. We inductively define x
≈
<y
1
,...,y
k
> as follows. x
≈
<y
1
,...,y
k+1
> if and only if
(
∃
z)(x
≈
<z,y
3
,...,y
k+1
>
∧
z
≈
<y
1
,y
2
>). In addition, we
define x
≈
<y> if and only if x = y.
LEMMA 5.7.3. Let k