Solution: Since S → ^ is
a production, S ⇒ ^. This implies that ^ ∈ L(G).
Now, for all n ≥ 1, we
can write the following:
S ⇒ 0S1 ⇒ 00S11
… ⇒ 0nS1n ⇒ 0n1n.
Therefore, 0n1n ∈ L(G).
In the above derivation, at every step, S → 0S1 is applied, except in the last step where S → ^ is applied.
Therefore, {0n1n n ≥ 0} ⊊ L(G).
Now suppose w ∈ L(G). So
we should start the derivation of w with S.
If we are applying
S → ^ first, then we will get w = ^.
Otherwise, the first
production that we need to apply is S → 0S1.
However, at any stage we can apply S → ^ to obtain the terminating string. Therefore
w can be derived in the following form.
S ⇒ 0nS1n � 0n1n,
for some n ≥ 1, That is L(G) ⊊ {0n1n | n ≥ 0}.
Hence L(G) = {0n1n n ≥ 0}.
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