Consider a context-free grammar 𝐺 with the following 3 rules. 𝑆 → 𝑎𝑆, 𝑆 →…

Key insight: count how many times each production is used.

• Let k be the number of times the rule aSbS is applied. Each such use contributes one b, so n_b = k. Each use of aSbS also increases the number of S nonterminals by 1; starting from one S there are k+1 final S's that must each produce a c, so n_c = k + 1.

• Let x be the number of times the rule aS is applied. Each aS adds one a without changing the number of S's, so n_a = x + k.

• From these counts we get n_c = n_b + 1, so the statement n_c(w) = n_b(w) + 1 holds for every w.

• Also n_a = x + k. Plugging into n_a(w) > n_c(w) − 2 gives x + k > (k + 1) − 2 ⇒ x > −1, which is always true since x ≥ 0. Hence n_a(w) > n_c(w) − 2 holds for every w.

• The statement n_a(w) > n_b(w) need not hold: n_a − n_b = x ≥ 0, so equality can occur (for example w = c or w = acbc). The statement n_c(w) = 2·n_b(w) is also not generally true (it would force k + 1 = 2k, i.e. k = 1 only).