Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 4J / c

Table of contents
- Solution c

Solution

In

H^{+}

the unstarred

S

occurs only as the initial symbol and has precisely the alternatives

S \to ε, S \to S^{*} .

The first produces only

ε

. After the second, the renamed-derivation argument from part (b) produces exactly

L (G)

. Therefore

L (H^{+}) = L (G) \cup {ε} .

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