Let us define a set S of binary strings according to the following rules:
Base: The empty string and the string 1 are in S.
Recursion: If xεS, then so are x0 and x11. (That is, x followed by 0 or 11.)
If S contains no other strings, which of the following strings are in S?
111,10101,11,1101,1011,010 or 00000
ϵ S, because 1 ϵ S and x ϵ S implies x11 ϵ S
not ϵ S, because if the string ends with 1 and its length is more
than 1, then it must end with 11.
ϵ S because the empty string is in S and x ϵ S → x11 ϵ S
e) 1011 ϵ S, because 1 ϵ S, 10 ϵ S, and 1011 ϵ S.
f) 010 not ϵ S. Assume the contrary. Then 01 ϵ S but (see the example b)). So, we get a contradiction q.e.d.
g) 00000 ϵ S because empty string is in S, therefore, 0 ϵ S, 00 ϵ S, 000 ϵ S etc.
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