Example 2:
S and E are mutually recursively defined hence the result must be
proved by a simultaneous induction on their rules.
For Rule S:The only option is that by the induction hypothesis, the E
tree has an even number of *, say m of them.
Then the *E* tree has M + 2 of the, which is an even value.
For Rule E:The 1st option builds a tree that has an even number of *,
because by the inductive hypothesis, the S tree has an even number,
and no new ones are added. The second option has exactly two
occurences, which is an even number.