Let:
\[ L1 = \{ a^n b^{2n} c^m \mid m, n \geq 1 \} \]
\[ L2 = \{ a^n b^m c^{2m} \mid m, n \geq 1 \} \]

a) Give CFGs for L1 and L2.

b) Is \( L1 \cap L2 \) a CFL? Justify your answer.

c) Using the CFG designed for L1 as a template, design another CFG for the language (denoted as \( L_{pref} \)) of all strings that are prefixes of the strings in L1 — i.e., ...

a) To give CFGs for L1 and L2, we can use the following rules:
For L1:
S -> aSc | T
T -> aTbb | ε

For L2:
S -> aSc | T
T -> aTbCc | ε

b) To determine if L1 T L2 is a CFL, we need to check if the intersection of the two languages is a CFL. We can see that the common substring in both languages is "a^n b^m", which is a regular language. Therefore, the intersection of L1 and L2 is a CFL since the intersection of a CFL and a regular language is always a CFL.

c) To design a CFG for Lpref, we can modify the CFG for L1 as follows:

S -> aSc | T | ε
T -> aTbb

The new rule S -> ε allows for the generation of empty prefixes, and the other rules remain the same as in the CFG for L1. This CFG generates all strings that are prefixes of the strings in L1.

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