CS 390 Solutions to Test 1
1. Prove
Let y be an arbitrary element of .
Then there is an element x in and y = f(x).
Hence and .
Hence and , that is . Hence .

2. For any set A, . Hence R is reflexive.
For two set A and B does not necessarily imply . Hence R is not symmetric.
For any three sets A,B, and C if and then . Hence R is transitive.

3.(a) .
(b)
(c) or
.

4. L is the set of strings consisting of a's and b's but containing at least one a.
Since the seed is a, any element of L is genmerated from a by applying (2) an appropriate number of times. Since both a and b can be appended any number of times any time during the generation and b can also be added in front any time any number of times, any string containing at least one a can be generated by (1) and (2).

5. The statement to be proven is as follows.
Any string in L starts with an a followed by any number of a's and then any number of b's.
Proof.
Basis Step:
Since the seed is a, we prove the statement for string a.
Obviously a starts with a and is followed by zero a's and zero b's.
Hence the statement holds for the basis case.
Induction Step:
Let x be a string in L.
Then we need to show that for any string produced from x by (2) of the definition, the statement holds.
(2) produces strings ax and xb from x.
For ax, since x starts with a and is followed by a's then b's, obviously ax still starts with a and is followed by a's and b's.
For xb, since x starts with a and is followed by a's and then b's, xb still starts with a, is followed by a's and then b's.
Hence the statement holds for all strings of L.

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