Mathematics for Computer Science

“mcs” — 2017/3/3 — 11:21 — page 209 — #217
7.2. Strings of Matched Brackets 209
Now,
[ ] [ ] D [ ] [ ] 2 RecMatch (letting s D ; t D [ ] )
[ [ ] ] D [ [ ] ] 2 RecMatch (letting s D [ ] ; t D )
[ [ ] ] [ ] 2 RecMatch (letting s D [ ] ; t D [ ] )
are also strings in RecMatch by repeated applications of the constructor case; and
so on.
It’s pretty obvious that in order for brackets to match, there had better be an equal
number of left and right ones. For further practice, let’s carefully prove this from
the recursive definitions, beginning with a recursive definition of the number # c .s/
of occurrences of the character c 2 A in a string s:
Definition 7.2.2.
Base case: # c ./ WWD 0.
Constructor case:
# c .ha; si/ WWD
(
# c .s/ if a ¤ c;
1 C # c .s/ if a D c:
The following Lemma follows directly by structural induction on Definition 7.2.2.
We’ll leave the proof for practice (Problem 7.9).
Lemma 7.2.3.
# c .s t/ D # c .s/ C # c .t/:
Lemma. Every string in RecMatch has an equal number of left and right brackets.
Proof. The proof is by structural induction with induction hypothesis
h
i
P.s/ WWD # [ .s/ D # ] .s/ :
Base case: P./ holds because
# [ ./ D 0 D # ] ./
by the base case of Definition 7.2.2 of # c ./.