The context free grammar for the language L = {an bm | n ≤ m + 3, n ≥ 0, m ≥…

Correct grammar: S → λ | aA | aaA | aaaA; A → aAb | B; B → Bb | λ

Why this works:

• For any given n and m with n ≤ m + 3 let k = min(n,m) and p = n − k (so 0 ≤ p ≤ 3).

• Use S → a^p A to produce the p unmatched leading a's (p is at most 3).

• Apply A → aAb exactly k times to produce a^k B b^k (these create k matched a/b pairs).

• Finally use B → Bb repeatedly (m − k times) to add the remaining b's, yielding a^{p+k} b^m = a^n b^m.

Conversely, any string derived from the grammar has at most three unmatched leading a's (provided by S), any further a's are paired with b's via A, and B only adds extra b's. Hence any derived string satisfies n − m ≤ 3, i.e. n ≤ m + 3.

Why the other given grammars fail:

• A grammar that forces exactly three leading a's (for example S → aaaA with no productions for fewer a's) cannot generate strings like 'a' or 'aa' or the empty string when those should be in the language.

• A grammar that omits the single-a or double-a start productions (S → aA or S → aaA) will miss valid strings where n = 1 or n = 2 and m is small.

Note: The solution supplied with the question was empty/absent. The grammar given above is the corrected complete solution and explains why it generates exactly the language L = { a^n b^m | n ≤ m + 3 }.