: [ S \to aSb \mid \varepsilon ]
: [ S \to aSbS \mid bSaS \mid \varepsilon ] cfg solved examples
: [ E \to E + T \mid T ] [ T \to T \times F \mid F ] [ F \to (E) \mid a \mid b ] : [ S \to aSb \mid \varepsilon ]
S ⇒ aSbb (first a) Now replace S with aSbb again? That would add another a. We need total 2 a’s. So second S must be ε: S ⇒ aSbb ⇒ a(aSbb)bb — now we have 2 a’s so S → ε: ⇒ a(aεbb)bb = aa b b b b = 2 a, 4 b (m=4). Not 3. 4 b (m=4). Not 3.