(a) The language $F = \setbuilder{a^i b^j c^k}{i,j,k \geq 0 \text{ and if } i = 1, \text{ then } j=k}$ is not regular.

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Proof: Suppose $F$ were regular. Let $G = \setbuilder{a b^i c^j}{i,j \geq 0}$. Observe that $G$ is regular. In particular, a regular expression for $G$ is $a b^* c^*$.

But, $F \cap G = H = \setbuilder{a b^n c^n}{n \geq 0}$. We show that $H$ is non-regular. Suppose it were and let $p$ be the pumping length. Take $z = a b^p c^p$, $z \in H$ and $|z| \geq p$. Take $z = uvw$, $|uv| \leq p$ and $|v| \geq 1$.

If $v = a$, then $u v^0 w = b^p c^p$ which is not in $H$. If $v = ab^k$ for some $1 \leq k < p$, then $u v^0 w = b^{(p-k)} c^p$ which is not in $H$. If $v = b^k$ for some $k \geq 1$, then $u v^0 w = a b^{(p-k)} c^p$ which is not in $H$. This contradicts the pumping lemma so $H$ is non-regular.

So, we have the the intersection of two regular languages producing a non-regular language, a contradiction. So it must be the case that $F$ is not regular.

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(b) $F$ satisfies the pumping lemma.

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Proof: Take the pumping length $p = 2$. We show that every string of length at least 2 can be pumped. Let $z \in F$ such that $|z| \geq 2$ be arbitrary.

Suppose there are no $a$'s in $z$. Then $z = b^i c^j$ with $i+j \geq 2$. Take $z = uvw$ where $u = \epsilon$, $v$ is the first symbol of $z$ and $w$ is the rest of $z$ (note $|uv| \leq 2$ and $|v| \geq 1$). Then clearly, $u v^m w \in F$ for any $k \geq 0$ because this simply changes the number of occurrences of the first symbol in $z$ (be it $b$ or $c$).

Suppose there is one $a$ in $z$. It must come first, so $z = a b^n c^n$, with $n \geq 1$. Take $z = uvw$ where $u = \epsilon$, $v = a$, and $w$ is the rest of $z$ (note $|uv| \leq 2$ and $|v| \geq 1$). Then $u v^m w \in F$ because this simply changes the number of $a$'s at the beginning of the string.

Suppose there are two $a$'s in $z$. Then $z = a^2 b^i c^j$ with $i,j \geq 0$. Take $z = uvw$ with $u = \epsilon$, $v = a^2$, and $w$ is the rest of $z$. Then $u v^m w \in F$ because this simply alters the amount of $a$'s in the string, but since it can never be the case that $v^m = a$, the restriction that the $b$'s and $c$'s must be equal in number is never encountered.

Suppose there are more than two $a$'s in $z$. Then $z = a^i b^j c^k$ with $i \geq 3$ and $j, k \geq 0$. Take $z = uvw$ with $u = \epsilon$, $v = a$, and $w$ is the rest of $z$. Then $u v^m w = a^y b^j c^k$ where $y \geq 2$ (that is, the lowest amount of $a$'s remaining after pumping is 2). Because $y \neq 1$, this is clearly in $F$.

So $F$ satisfies the pumping lemma.