A machine is represented by states Q, input alphabet Σ, transition function δ,…

Given language description (interpreted clearly)

The machine accepts all strings over Σ={a,b}\Sigma = \{a, b\}Σ={a,b} such that:

• The string may start and end with any combination of symbols from {a,b}\{a,b\}{a,b}, and

• The substring abb must occur somewhere in the string.

In formal language terms, the language is:

L=Σ∗ abb Σ∗L = \Sigma^*\,abb\,\Sigma^*L=Σ∗abbΣ∗

Given grammar

S→A abb AA→aA∣bA∣ε\begin{aligned} S &\rightarrow A\,abb\,A \\ A &\rightarrow aA \mid bA \mid \varepsilon \end{aligned}SA​→AabbA→aA∣bA∣ε​

Verification of the grammar

• Non-terminal AAA generates:

  A⇒Σ∗(any string over {a,b},including empty string)A \Rightarrow \Sigma^* \quad (\text{any string over } \{a,b\}, \text{including empty string})A⇒Σ∗(any string over {a,b},including empty string)

• Start symbol SSS generates:

  S⇒Σ∗ abb Σ∗S \Rightarrow \Sigma^*\,abb\,\Sigma^*S⇒Σ∗abbΣ∗

This exactly matches the language description:

• Any prefix of aaa's and bbb's

• Followed by the fixed substring abb

• Followed by any suffix of aaa's and bbb's

✅ Conclusion

✔ The given grammar correctly represents the language accepted by the machine.

S→AabbA,A→aA∣bA∣ε\boxed{ S \rightarrow AabbA,\quad A \rightarrow aA \mid bA \mid \varepsilon }S→AabbA,A→aA∣bA∣ε​

This grammar generates all and only those strings over {a,b}\{a,b\}{a,b} that contain abb as a substring, regardless of what appears before or after it.