Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 4H / b / iii

Table of contents
- Solution iii

Solution

3 m = 4 l

and

2 n = 5 k

are equivalent to

m = 4 r, l = 3 r, n = 5 s, k = 2 s

for nonnegative integers

r, s

. Hence the language is

{a^{4 r} b^{5 s} c^{2 s} d^{3 r} : r, s \geq 0},

which is the nested two-pair count language generated by

S \to aaaa S ddd ∣ T, T \to bbbbb T cc ∣ ϵ .

It is therefore context-free.

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