Consider the following grammar: \(𝑆→𝑋𝑌 \\ 𝑋→𝑌𝑎𝑌∣𝑎 \ \ and \ \…

Answer: Every string produced by the grammar has at least two a's and has a number of b's that is a multiple of 2.

Reasoning (b's are in multiples of 2):

• The production for Y is Y → bbX, so each time a Y expands it contributes exactly two b's followed by whatever X produces.

• X can be a (which contributes zero b's) or X → Y a Y, which uses two Y's; since each Y contributes an even number (two plus the b's from an X), the total number of b's contributed by X is always even.

• Thus every derivation produces an even number of b's, i.e. b's occur in multiples of 2.

Reasoning (at least two a's):

• Start symbol is S → X Y. Any terminal string from X must contain at least one a because X can reduce directly to a, and any other expansion of X (X → Y a Y) explicitly contains an a.

• The Y in S expands to bbX, which introduces another X that again must yield at least one a.

• Therefore every complete string from S contains at least two occurrences of a.

Why the other statements are false:

• Three consecutive a's cannot appear because any a coming from a Y expansion is separated from other terminals by the "bb" that Y produces; there is no production that places two X-derived a's adjacent without intervening b's.

• Strings do not have strictly alternating a and b because Y introduces "bb", so there are consecutive b's in typical strings (for example: a b b a).