where P
η
is the sum over the admissible lines of the diagram P b
N −1
, b
N
; ˜ C
N −2
. Using these inequalities and Lemma 5.6, the sum over b
N
in the second line of 6.72 is bounded as X
b
N
p
b
N
M
1
b
N −1
,b
N
;˜ C
N −2
P b
N
, b
N +1
; ˜ C
N −1
, Bu, ℓ~x
I
1
{a∈˜C
N −1
}
≤ X
η
X
c
X
b
N
P b
N −1
, b
N
; ˜ C
N −2
, ℓ
η
c, ℓa
p
b
N
P b
N
, b
N +1
; c, Bu, ℓ~x
I
+ P b
N −1
, b
N
; ˜ C
N −2
, ℓ
η
c p
b
N
P b
N
, b
N +1
; c, Bu, ℓ~x
I
, ℓa
≤ P
1
b
N −1
, b
N +1
; ˜ C
N −2
, Bu, ℓ~x
I
, ℓa
, 6.81
where we have used the fact that the rightmost expression has more possibilities for the lines on which Construction
ℓa can be performed, as in the proof of Lemma 5.6. Finally, by a version of
6.14, we obtain 6.72
≤ X
I ⊂J
X
a ,u
t
a
≥t
bN+1
P
1
b
N −1
, b
N +1
; ˜ C
N −2
, Bu, ℓa, ℓ~x
I
P
u , y
2
; a, ℓ~x
J \I
≤ P
1
b
N −1
, b
N +1
; ˜ C
N −2
, E
t
bN+1
y
2
, ℓ~x
. 6.82
This completes the proof of 6.64 assuming 6.76. It remains to show 6.76. By definition, there is a line, say,
τw − v for some w with t
w
≤ t
a
, contained in the diagram function S
0,1
b
N
, v ; c, 2
v
b
N +1
, Bu, ℓ~x
I
. We claim that
τw − v τa − v ≤ τ w − v; ℓa ,
6.83 which readily implies 6.76. To show 6.83, we let P
1
, P
2
be independent percolation measures and denote by P
1,2
their product measure. Then, we can rewrite the left-hand side of 6.83 as
τw − v τa − v = P
1,2
[
γ
1
:v →w
γ
2
:v →a
n γ
1
is 1-occupied, γ
2
is 2-occupied o
. 6.84
Taking note of the last common vertex between γ
1
and γ
2
and using the Markov property, we obtain 6.84
≤ X
z
τz − v
2
P
1,2
[
γ
′ 1
:z →w
γ
′ 2
:z →a
γ
′ 1
∩γ
′ 2
={z}
n γ
′ 1
is 1-occupied, γ
′ 2
is 2-occupied o
. 6.85
If z = w , then the above probability P
1,2
· · · equals τa− w . If z 6= w hence z 6= a, then at least
one of γ
′ 1
and γ
′ 2
has to leave z with a spatial bond. Recalling the definition of Construction Bz
and applying the naive inequality τz − v
2
≤ τz − v to 6.85, we conclude
6.85 ≤
X
z
′
τ w − v; Bz
′
τa − z
′
≡ τ w − v; ℓa
. 6.86
This completes the proof of 6.83, hence the proof of 6.76. 870
Proof of Lemma 6.7 for N
1
≥ 1. First we recall that, by 6.3 and 5.40, B
N1
δ
b
N +1
, y
1
; ˜ C
N
≤ X
b= · ,y
1
P
N1−1
b
N +1
, b; ˜ C
N
p
b
, 6.87
where, by 5.39,
P
N1−1
b
N +1
, b; ˜ C
N
= P
b
N +1
, b; ˜ C
N
N
1
= 1, ≤
X
η
X
z
X
e
P b
N +1
, e; ˜ C
N
, ℓ
η
z p
e
P
N1−2
e, b; z N
1
≥ 2. 6.88
Then, by following the argument between 6.72 and 6.82 and using versions of 6.57 and 5.59, we obtain that, for N
1
≥ 2, X
b
N +1
p
b
N +1
M
N +1
b
N +1
1
V
t y1 −ǫ
b
N
,y
2
∩ {~x ∈˜C
N
}
P b
N +1
, e; ˜ C
N
, ℓ
η
z
≤ X
b
N +1
X
η
′
X
c
P
N
b
N +1
; ℓ
η
′
c, V
t
y 1
−ǫ
y
2
, ℓ~x
p
b
N +1
P b
N +1
, e; c,
ℓ
η
z
≤ P
N +1
e; V
t
y 1
−ǫ
y
2
,
ℓ~x , ℓ
η
z = R
N +1
e, y
2
;
ℓ~x , ℓ
η
z .
6.89 For N
1
= 1, we simply ignore ℓ
η
z and replace e by b, which immediately yields 6.45. For N
1
≥ 2, by a version of 5.59, we obtain
LHS of 6.45 ≤
X
b= · ,y
1
X
η
X
z
X
e
R
N +1
e, y
2
; ℓ~x , ℓ
η
z p
e
P
N1−2
e, b; z p
