Additive Combinatorics
and its Applications in
Theoretical CS
By Shachar Lovett
Chapter 2: Set addition, Sections 2.1-2.3, pp. 3-8.
Presented by Tomer Bincovich
Set addition
• 𝐺 an abelian group (think about 𝐺 = 𝔽𝑛 ).
• 𝐴 ⊆ 𝐺.
• The sumset of 𝐴 is
2𝐴 = 𝐴 + 𝐴 = 𝑎 + 𝑎′ 𝑎, 𝑎′ ∈ 𝐴
• Always 2𝐴 ≥ 𝐴 .
• When does equality hold?
When does 2𝐴 = |𝐴|?
• Equality holds if 𝐴 is empty, a subgroup of 𝐺 or a coset of a subgroup.
• For the other direction:
• If 𝐴 ≠ ∅, we can assume w.l.o.g. 0 ∈ 𝐴 by shifting.
• Then 𝐴 ⊆ 2𝐴, and since 2𝐴 = 𝐴 we have that 2𝐴 = 𝐴.
• 𝐴 is a nonempty finite subset that is closed under addition, hence is a
subgroup.
Planned topics
• Ruzsa calculus
• The span of sets of small doubling
• The growth of sets of small doubling
Ruzsa calculus
• A set of basic inequalities between sizes of sets and their sumsets.
• For 𝐴, 𝐵 ⊆ 𝐺, define:
• Sumset: 𝐴 + 𝐵 = 𝑎 + 𝑏 𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵
• Difference set: 𝐴 − 𝐵 = 𝑎 − 𝑏 𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵
Claim 2.1 (Ruzsa triangle inequality):
𝐴 𝐵−𝐶 ≤ 𝐴−𝐵 𝐴−𝐶
Proof of Claim 2.1
• Claim 2.1 (Ruzsa triangle inequality):
𝐴 𝐵−𝐶 ≤ 𝐴−𝐵 𝐴−𝐶
• Define a map 𝑓: 𝐴 × 𝐵 − 𝐶 → 𝐴 − 𝐵 × 𝐴 − 𝐶 :
for any 𝑥 ∈ 𝐵 − 𝐶 fix 𝑏 ∈ 𝐵, 𝑐 ∈ 𝐶 s.t. 𝑥 = 𝑏 − 𝑐, and define
𝑓 𝑎, 𝑥 = 𝑎 − 𝑏, 𝑎 − 𝑐 .
• 𝑓 is injective, hence 𝐴 𝐵 − 𝐶 ≤ 𝐴 − 𝐵 𝐴 − 𝐶 .
The Ruzsa distance
𝐴−𝐵
𝑑 𝐴, 𝐵 = log 1 2 1
𝐴
𝐵
• Not formally a distance function. Why?
2
• Claim 2.3: The Ruzsa distance is symmetric and obeys the triangle
inequality.
Proof of Claim 2.3
𝐴−𝐵
𝑑 𝐴, 𝐵 = log 1 2 1
𝐴
𝐵
• Symmetry: since 𝐵 − 𝐴 = 𝐴 − 𝐵 .
• Moreover, 𝑑 𝐴, 𝐶 ≤ 𝑑 𝐴, 𝐵 + 𝑑 𝐵, 𝐶 is equivalent to
𝐴−𝐶
𝐴−𝐵
𝐵−𝐶
log 1 2 1 2 ≤ log 1 2 1 2 + log 1 2 1
𝐴
𝐶
𝐴
𝐵
𝐵
𝐶
𝐴−𝐶
𝐴1 2𝐶1
2
𝐴−𝐵
≤
𝐴1 2𝐵1
2
𝐵−𝐶
𝐵1 2𝐶1
𝐵 𝐴−𝐶 ≤ 𝐵−𝐴 𝐵−𝐶
Which follows from Claim 2.1.
2
2
2
Corollary 2.4
If 𝐴 − 𝐵 ≤ 𝐾 𝐴
1 2
𝐵
1 2
then 𝐴 − 𝐴 ≤ 𝐾 2 𝐴 .
• For 𝐵 = −𝐴 we get that if 𝐴 + 𝐴 ≤ 𝐾 𝐴 then 𝐴 − 𝐴 ≤ 𝐾 2 𝐴 .
Proof of Corollary 2.4
If 𝐴 − 𝐵 ≤ 𝐾 𝐴
1 2
𝐴−𝐴
= 𝑒𝑑
𝐴
𝐵
𝐴,𝐴
1 2
≤𝑒
then 𝐴 − 𝐴 ≤ 𝐾 2 𝐴 .
𝑑 𝐴,𝐵 +𝑑 𝐵,𝐴
=
𝐴−𝐵
𝐴1 2𝐵1
𝐴−𝐵
𝑑 𝐴, 𝐵 = log 1 2 1
𝐴
𝐵
2
2
2
≤ 𝐾2
The doubling constant
• The doubling constant of 𝐴 is 𝐾 =
𝐴+𝐴
𝐴
.
• We saw that 𝐾 = 1 corresponds to subgroups.
• Can we use 𝐾 to say something about the size of the smallest
subgroup containing 𝐴? (or, in the case of 𝐺 = 𝔽𝑛 , the size of the
span of 𝐴)
Lemma 2.5 (Laba)
If 𝐴 − 𝐴 <
3
2
𝐴 then 𝐴 − 𝐴 is a subgroup.
• The constant 3/2 is tight: take 𝐴 = 0,1 ⊆ ℤ.
𝐴 − 𝐴 = −1,0,1 is not a subgroup.
Proof of Lemma 2.5
• We will show that for any 𝑥 ∈ 𝐴 − 𝐴, 𝐴 ∩ 𝐴 + 𝑥 > 𝐴 /2.
• This implies that for any 𝑥, 𝑦 ∈ 𝐴 − 𝐴, 𝐴 + 𝑥 ∩ 𝐴 + 𝑦 ≠ ∅:
• Denoting 𝐴𝑥 = 𝐴 ∩ 𝐴 + 𝑥 , by the inclusion exclusion principle
𝐴+𝑥 ∩ 𝐴+𝑦
≥ 𝐴𝑥 ∩ 𝐴𝑦 = 𝐴𝑥 + 𝐴 𝑦 − 𝐴𝑥 ∪ A 𝑦
> 𝐴 /2 + 𝐴 /2 − 𝐴 = 0
• There are 𝑎1 , 𝑎2 ∈ 𝐴 s.t. 𝑎1 + 𝑥 = 𝑎2 + 𝑦, thus 𝑥 − 𝑦 = 𝑎2 − 𝑎1
∈ 𝐴 − 𝐴.
• 𝐴 − 𝐴 is closed under taking difference, hence must be a subgroup.
Proof of Lemma 2.5 cont.
• We will show that for any 𝑥 ∈ 𝐴 − 𝐴, 𝐴 ∩ 𝐴 + 𝑥 > 𝐴 /2.
• Let 𝑥 = 𝑎 − 𝑎′ ∈ 𝐴 − 𝐴. Then 𝐴 ∩ 𝐴 + 𝑥 = 𝐴 − 𝑎 ∩ 𝐴 − 𝑎′ .
• Similarly to before:
𝐴 − 𝑎 ∩ 𝐴 − 𝑎′
= 𝐴 − 𝑎 + 𝐴 − 𝑎′ − 𝐴 − 𝑎 ∪ 𝐴 − 𝑎′
≥ 𝐴 + 𝐴 − 𝐴 − 𝐴 > |𝐴|/2
As needed.
Lemma 2.6 (Freiman)
Let 𝐴 ⊆ ℝ𝑛 with 𝐴 + 𝐴 ≤ 𝐾 𝐴 . Then 𝐴 lies in an affine subspace of
dimension at most 2𝐾 − 1.
Proof of Lemma 2.6
• We will prove that if 𝐴 has affine dimension 𝑑 then
𝑑+1
𝐴+𝐴 ≥ 𝑑+1 𝐴 −
2
• Thus 𝑑 + 1 𝐴 −
𝑑+1
2
≤ 𝐴 + 𝐴 ≤ 𝐾 𝐴 , and since 𝐴 ≥ 𝑑,
𝑑+1
𝑑+1
𝑑+1−𝐾 𝑑 ≤ 𝑑+1−𝐾 𝐴 ≤
=
𝑑
2
2
• Hence 𝐾 ≥
𝑑+1
,
2
or 𝑑 ≤ 2𝐾 − 1.
Proof of Lemma 2.6 cont.
• We prove by induction on 𝐴 that if 𝐴 has affine dimension 𝑑 then
𝑑+1
𝐴+𝐴 ≥ 𝑑+1 𝐴 −
2
• If 𝐴 = 1 then 𝐴 + 𝐴 = 1, 𝑑 = 0 and the claim follows.
𝑎+𝑎′
2
• Assume 𝐴 > 1. Let 𝑀 𝐴 =
𝑎, 𝑎′ ∈ 𝐴 .
Clearly 𝑀 𝐴 = 𝐴 + 𝐴 .
• Let 𝐶(𝐴) be the convex hull of 𝐴, which is a 𝑑-dimensional polytope
by assumption. Its vertices are the points in 𝐴 that cannot be
obtained as a convex combination of the other points.
Inductive proof, case (1)
• Fix a vertex 𝑎∗ ∈ 𝐴 of 𝐶 𝐴 , and let 𝐴′ = 𝐴 ∖ 𝑎∗ . We have two cases:
𝑎+𝑎∗
2
′
1) The affine dimension of 𝐴 is 𝑑 − 1. Then the mid-points
for all
𝑎 ∈ 𝐴 are outside 𝐶(𝐴′ ).
𝑑
𝑑
′
′
𝑀 𝐴 ≥ 𝐴 + 𝑀 𝐴 ≥ 𝐴 +𝑑 𝐴 −
= 𝑑+1 𝐴 −𝑑−
2
2
𝑑+1
= 𝑑+1 𝐴 −
2
?
𝑑+1
𝐴+𝐴 ≥ 𝑑+1 𝐴 −
2
Inductive proof, case (2)
• Fix a vertex 𝑎∗ ∈ 𝐴 of 𝐶 𝐴 , and let 𝐴′ = 𝐴 ∖ 𝑎∗ . We have two cases:
2) The affine dimension of 𝐴′ is 𝑑. Let 𝑎1 , … , 𝑎𝑡 be the neighboring vertices
to 𝑎∗ in 𝐶 𝐴 (two vertices are neighbors if the segment connecting
them is a one-dim. face of 𝐶 𝐴 ). Note that 𝑡 ≥ 𝑑 since 𝐶 𝐴 is 𝑑-dim.
𝑎∗ +𝑎𝑖
∗
The points 𝑎 and
for 1 ≤ 𝑖 ≤ 𝑡 are mid-points outside 𝐶 𝐴′ .
2
𝑑+1
′
′
𝑀 𝐴 ≥ 𝑡+1 + 𝑀 𝐴 ≥𝑑+1+ 𝑑+1 𝐴 −
2
𝑑+1
= 𝑑+1 𝐴 −
?
𝑑+1
2
𝐴+𝐴 ≥ 𝑑+1 𝐴 −
2
Lemma 2.6 – tightness
Let 𝐴 ⊆ ℝ𝑛 with 𝐴 + 𝐴 ≤ 𝐾 𝐴 . Then 𝐴 lies in an affine subspace of
dimension at most 2𝐾 − 1.
• The dimension is tight up to lower order terms. Take 𝐴 = 𝑣1 , … , 𝑣2𝐾
a set of linearly independent vectors, then 𝐴 + 𝐴 = 2𝐾
+ 2𝐾
2
𝐴+𝐴
1
= 𝐾(2𝐾 + 1) and
=𝐾+ .
𝐴
2
Theorem 2.7 (Ruzsa)
Let 𝐺 be an Abelian group of torsion 𝑟, 𝐴 ⊆ 𝐺 with 𝐴 + 𝐴 ≤ 𝐾 𝐴 .
Then there exists a subgroup 𝐻 < 𝐺, 𝐻 ≤
4
2
𝐾
𝐾 𝑟
𝐴 s.t. 𝐴 ⊆ 𝐻.
• A group has torsion 𝑟 ≥ 1 if 𝑟 ∙ 𝑔 = 0 for all 𝑔 ∈ 𝐺.
Example
• 𝐺 = 𝔽𝑛𝑝 has torsion 𝑟 = 𝑝. Let 𝑈, 𝑉 ⊆ 𝔽𝑛𝑝 two subspaces with
𝑈 ∩ 𝑉 = 0 , and set 𝐴 = 𝑈 + {𝑣1 , … , 𝑣2𝐾 } where 𝑣1 , … , 𝑣2𝐾 ∈ 𝑉 are
linearly independent.
• 𝐴 = 𝑈 ∙ 2𝐾. Since 𝐴 + 𝐴 = 𝑈 + 𝑣𝑖 + 𝑣𝑗 1 ≤ 𝑖 ≤ 𝑗 ≤ 2𝐾 ,
𝐴+𝐴
1
𝐴 + 𝐴 = 𝑈 ∙ 𝐾 2𝐾 + 1 and
= 𝐾 + ≈ 𝐾.
𝐴
2
• However, the size of the minimal subspace containing 𝐴 is
2𝐾
𝑝
𝑝2𝐾 𝑈 =
𝐴
2𝐾
• Thus an exponential dependency on 𝐾 is unavoidable.
Conjecture 2.8 (Ruzsa)
There exists an absolute constant 𝐶 ≥ 2 s.t. for any Abelian group 𝐺 of
torsion 𝑟, and any 𝐴 ⊆ 𝐺 with 𝐴 + 𝐴 ≤ 𝐾|𝐴|, the subgroup generated
by 𝐴 has order ≤ 𝑟 𝐶𝐾 𝐴 .
• Was established with 𝐶 = 2 for 𝑟 = 2 and then extended for prime 𝑟:
• Theorem 2.9:
Let 𝑝 be a prime, 𝐴 ⊆ 𝔽𝑛𝑝 with 𝐴 + 𝐴 ≤ 𝐾 𝐴 . Then there exists a
subspace 𝐻 ⊆ 𝔽𝑛𝑝 s.t. 𝐴 ⊆ 𝐻 and 𝐻 ≤
𝑝2𝐾
2𝐾−1
𝐴.
Theorem 2.10
If 2𝐴 ≤ 𝐾 𝐴 then 2𝐴 − 2𝐴 ≤ 𝐾 4 𝐴 .
• Used to prove Theorem 2.7.
• We will prove a more general result later today.
Proof of Theorem 2.7 - Sketch
• Let 𝐴 ⊆ 𝐺 with 2𝐴 ≤ 𝐾 𝐴 . We can assume 0 ∈ 𝐴 by possibly
replacing 𝐴 with 𝐴 − 𝑎 for some 𝑎 ∈ 𝐴.
• Let 𝐵 = 𝑏1 , … , 𝑏𝑚 ⊆ 2𝐴 − 𝐴 be a maximal collection of elements
s.t. 𝑏𝑖 − 𝐴 are all disjoint (“Ruzsa covering”).
• 𝐴 has small doubling, thus Theorem 2.10 implies that 𝐵 is bounded.
• We will show that ℓ𝐴 − 𝐴 ⊆ ℓ − 1 𝐵 + 𝐴 − 𝐴 for all ℓ ≥ 1.
• Together the theorem follows.
Proof of Theorem 2.7 cont.
• Since 𝑏𝑖 − 𝐴 ⊆ 2𝐴 − 2𝐴,
2𝐴 − 2𝐴
𝑚≤
≤ 𝐾4
𝐴
• Therefore 𝐵 ≤ 𝐾 4 .
2𝐴 ≤ 𝐾 𝐴 ⇒ 2𝐴 − 2𝐴 ≤ 𝐾 4 𝐴
ℓ𝐴 − 𝐴 ⊆ ℓ − 1 𝐵 + 𝐴 − 𝐴
• ℓ = 1 is trivial.
• For ℓ = 2, we need to show 2𝐴 − 𝐴 ⊆ 𝐵 + 𝐴 − 𝐴.
• Take 𝑥 ∈ 2𝐴 − 𝐴, by construction there exists 𝑏 ∈ 𝐵 s.t.
𝑥 − 𝐴 ∩ 𝑏 − 𝐴 ≠ ∅, i.e. 𝑥 − 𝑎 = 𝑏 − 𝑎′ for some 𝑎, 𝑎′ ∈ 𝐴.
• Hence 𝑥 = 𝑏 + 𝑎 − 𝑎′ ∈ 𝐵 + 𝐴 − 𝐴.
• By induction on ℓ,
ℓ𝐴 − 𝐴 = 𝐴 + ℓ − 1 𝐴 − 𝐴 ⊆ 𝐴 + ℓ − 2 𝐵 + 𝐴 − 𝐴
⊆ ℓ−2 𝐵+𝐵+𝐴−𝐴 = ℓ−1 𝐵+𝐴−𝐴
Concluding the proof
ℓ𝐴 − 𝐴 ⊆ ℓ − 1 𝐵 + 𝐴 − 𝐴
• Let 𝐴 be the subgroup spanned by 𝐴, and similarly for 𝐵.
𝐴 =
ℓ𝐴 ⊆
ℓ≥1
ℓ𝐴 − 𝐴 ⊆
ℓ≥1
ℓ−1 𝐵+𝐴−𝐴 = 𝐵 +𝐴−𝐴
ℓ≥1
• Hence, using Corollary 2.4 (which implies 𝐴 − 𝐴 ≤ 𝐾 2 𝐴 ),
𝐴 ≤ 𝐵 ∙ 𝐴 − 𝐴 ≤ 𝐵 ∙ 𝐾2 𝐴
• We conclude by bounding 𝐵 . As 𝐺 has torsion 𝑟 and we saw that
4
4
𝐾
𝐵 ≤ 𝐾 we have 𝐵 ≤ 𝑟 .
• Finally, 𝐴 ≤
4
2
𝐾
𝐾 𝑟
𝐴.
Iterated sumsets
ℓ𝐴 = 𝑎1 + ⋯ + 𝑎ℓ 𝑎1 , … , 𝑎ℓ ∈ 𝐴
ℓ𝐴 − 𝑚𝐴 = 𝑎1 + ⋯ + 𝑎ℓ − 𝑎ℓ+1 − ⋯ − 𝑎ℓ+𝑚 𝑎1 , … , 𝑎ℓ+𝑚 ∈ 𝐴
• Theorem 2.11 (Plünneke-Ruzsa): If 𝐴 = 𝐵 and 𝐴 + 𝐵 ≤ 𝐾 𝐴
then ℓ𝐴 − 𝑚𝐴 ≤ 𝐾 ℓ+𝑚 𝐴 .
• For 𝐵 = 𝐴 or 𝐵 = −𝐴 we obtain the following:
• Corollary 2.12: If 𝐴 + 𝐴 ≤ 𝐾 𝐴 or 𝐴 − 𝐴 ≤ 𝐾 𝐴
then ℓ𝐴 − 𝑚𝐴 ≤ 𝐾 ℓ+𝑚 𝐴 .
Lemma 2.13
Let 𝐴, 𝐵 ⊆ 𝐺 with 𝐴 = 𝐵 and 𝐴 + 𝐵 ≤ 𝐾 𝐴 . Let 𝐵0 ⊆ 𝐵 a
𝐴+𝐵0
nonempty set minimizing the ratio 𝐾0 =
. Then for any 𝐶 ⊆ 𝐺,
𝐵0
𝐴 + 𝐵0 + 𝐶 ≤ 𝐾0 𝐵0 + 𝐶
• Notice that 𝐾0 ≤ 𝐾.
Proof of Theorem 2.11
•
•
•
•
Let 𝐵0 ⊆ 𝐵 as in Lemma 2.13. ∀𝐶 𝐴 + 𝐵0 + 𝐶 ≤ 𝐾0 𝐵0 + 𝐶
We will prove by induction on ℓ that 𝐵0 + ℓ𝐴 ≤ 𝐾0ℓ 𝐵0 ≤ 𝐾 ℓ 𝐵0 .
ℓ = 0 is easy: 𝐵0 + 0𝐴 = 1 ∙ 𝐵0 .
For ℓ > 0:
𝐵0 + ℓ𝐴 = 𝐴 + 𝐵0 + (ℓ − 1)𝐴 ≤ 𝐾0 𝐵0 + ℓ − 1 𝐴
≤ 𝐾0 ∙ 𝐾0ℓ−1 𝐵0 = 𝐾0ℓ 𝐵0
• By the Ruzsa triangle inequality (Claim 2.1), applied to −𝐵0 , ℓ𝐴, 𝑚𝐴:
−𝐵0 ℓ𝐴 − 𝑚𝐴 ≤ 𝐵0 + ℓ𝐴 𝐵0 + 𝑚𝐴 ≤ 𝐾 ℓ+𝑚 𝐵0 2
𝐴 𝐵−𝐶 ≤ 𝐴−𝐵 𝐴−𝐶
• Hence
ℓ𝐴 − 𝑚𝐴 ≤ 𝐾 ℓ+𝑚 𝐵0 ≤ 𝐾 ℓ+𝑚 𝐴
Lemma 2.13 (Reminder)
Let 𝐴, 𝐵 ⊆ 𝐺 with 𝐴 = 𝐵 and 𝐴 + 𝐵 ≤ 𝐾 𝐴 . Let 𝐵0 ⊆ 𝐵 a
𝐴+𝐵0
nonempty set minimizing the ratio 𝐾0 =
. Then for any 𝐶 ⊆ 𝐺,
𝐵0
𝐴 + 𝐵0 + 𝐶 ≤ 𝐾0 𝐵0 + 𝐶
Proof of Lemma 2.13
• Observe that by minimality of 𝐾0 , for any 𝐵′ ⊆ 𝐵 we have that
𝐴 + 𝐵′ ≥ 𝐾0 𝐵′ .
• The proof is by induction on 𝐶 .
• If 𝐶 = 𝑥 ,
𝐴 + 𝐵0 + 𝑥 = 𝐴 + 𝐵0 = 𝐾0 𝐵0 = 𝐾0 𝐵0 + 𝑥
• If 𝐶 > 1, write 𝐶 = 𝐶 ′ ∪ 𝑥 .
• Define 𝐵′ ⊆ 𝐵0 as 𝐵′ = 𝑏 ∈ 𝐵0 𝐴 + 𝑏 + 𝑥 ⊆ 𝐴 + 𝐵0 + 𝐶 ′ .
• We can write
𝐴 + 𝐵0 + 𝐶 = 𝐴 + 𝐵0 + 𝐶 ′ ∪ 𝐴 + 𝐵0 + 𝑥 ∖ 𝐴 + 𝐵′ + 𝑥
Proof of Lemma 2.13 cont.
𝐴 + 𝐵0 + 𝐶 = 𝐴 + 𝐵0 + 𝐶 ′ ∪
𝐴 + 𝐵0 + 𝑥 ∖ 𝐴 + 𝐵′ + 𝑥
𝐴 + 𝐵0 + 𝐶 ≤ 𝐴 + 𝐵0 + 𝐶 ′ + 𝐴 + 𝐵0 + 𝑥 ∖ 𝐴 + 𝐵′ + 𝑥
= 𝐴 + 𝐵0 + 𝐶 ′ + 𝐴 + 𝐵0 + 𝑥 − 𝐴 + 𝐵′ + 𝑥
≤ 𝐾0 𝐵0 + 𝐶 ′ + 𝐴 + 𝐵0 − 𝐴 + 𝐵′ ∀𝐵′ ⊆ 𝐵
′ ≥ 𝐾 𝐵′
′
′
𝐴
+
𝐵
0
≤ 𝐾0 𝐵0 + 𝐶 + 𝐾0 𝐵0 − 𝐾0 𝐵
= 𝐾0 𝐵0 + 𝐶 ′ + 𝐵0 − 𝐵′
• It remains to show that:
𝐵0 + 𝐶 ′ + 𝐵0 − 𝐵′ ≤ 𝐵0 + 𝐶
Concluding the proof
𝐵′ = 𝑏 ∈ 𝐵0 𝐴 + 𝑏 + 𝑥 ⊆ 𝐴 + 𝐵0 + 𝐶 ′
• It remains to show that:
𝐵0 + 𝐶 ′ + 𝐵0 − 𝐵′ ≤ 𝐵0 + 𝐶
• Define 𝐵′′ = 𝑏 ∈ 𝐵0 𝑏 + 𝑥 ∈ 𝐵0 + 𝐶 ′ . Note that 𝐵′′ ⊆ 𝐵′ .
• We decompose 𝐵0 + 𝐶 as a disjoint union:
𝐵0 + 𝐶 = 𝐵0 + 𝐶 ′ ∪ 𝐵0 + 𝑥 ∖ 𝐵′′ + 𝑥
• Thus
𝐵0 + 𝐶 = 𝐵0 + 𝐶 ′ + 𝐵0 − 𝐵′′ ≥ 𝐵0 + 𝐶 ′ + 𝐵0 − 𝐵′
Which is what we needed to show.