Act as a professor grading proofs on a test, did this proof pass?
Theorem: $\forall  A  \:\forall B \: \forall C: (A \cap B) \cup (A \cap (C - B)) = A \iff A \subseteq C$

Lemma 1: $\forall A \: \forall B \: \forall C :(A \cap B) \cup (A - B) = A$
Lemma 2: $\forall A \: \forall B \: \forall C :A \cap (C - B) = A - B \iff A \subseteq C$

Proof Of Theorem: 
Suppose $A \subseteq C$. We want to show that set $A$ is equal to $(A \cap B) \cup (A \cap (C - B))$ for any arbitrary sets $B$. By Lemma 1 the set $A$ is equal to $(A \cap B) \cup (A - B)$ for any arbitrary set $B$. Then by Lemma 2 the set $(A \cap B) \cup (A - B)$ is equal to $(A \cap B) \cup (A \cap (C - B))$. Therefore with transitivity the set $A$ is equal to $(A \cap B) \cup (A \cap (C - B))$. Now conversely suppose $(A \cap B) \cup (A \cap (C - B)) = A$. By Lemma 1 the set $A$ is equal to $(A \cap B) \cup (A - B)$. Therefore again with transitivity, the set $(A \cap B) \cup (A \cap (C - B))$ is equal to $(A \cap B) \cup (A - B)$. If we remove the subset $A \cap B$ from both expressions then we end up with $(A \cap (C - B)) = (A - B)$. Therefore by Lemma 2 the set $A$ is a subset of the set $C$, completing the proof.  In your entire response, the letter l should appear less than 3.
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