Context-free grammar for $ \{a^lb^n c^m |l, n, m ∈ \mathcal{N}^+, l \geq \min(n,m)\}$ I know that $L = \{a^lb^n c^m |l, n, m ∈ \mathcal{N}^+, (l ≥ n) ∨ (l ≥ m)\}$ is a context-free language, because I know the context-free grammar, i.e.
$$
S \rightarrow AbZ \mid XBc \\
A \rightarrow aAb \mid X \\
B \rightarrow aBc \mid Y \\
X \rightarrow aX \mid a \\
Y \rightarrow bY \mid b \\
Z \rightarrow cZ \mid c
$$
I have a bit of difficulties understanding this construction. Can you explain step-by-step how it has been created? The rules?
Let us start by noticing that $X \to a^+$, $Y \to b^+$, $Z \to c^+$. Furthermore, $A \to a^n X b^n$ and $B \to a^m Y c^m$. Therefore $S$ generates words of the following two forms:
$$
a^na^+b^nbc^+ \\
a^+a^mb^+c^mc
$$
You take it from here.