00:02
We're asked to prove that s is 0 related to t if and only if s and t are both final states, or neither are final states.
00:19
So, first, let s be 0 related to t.
00:30
This is part a, by the way.
00:38
This implies that f of s x and if tx are both final or non -final.
00:57
States for all strings x such that the length of x is less than or equal to zero.
01:15
In other words, since lambda, the empty string is the only string of length zero, it follows that f of s lambda and f of t lambda are either both final or both non -final states, where lambda is the empty string.
01:43
Now we have that by definition f of s lambda is s and f of t lambda is t.
02:02
So we have that s and t are both final states or both non -final states.
02:18
Now that was one direction.
02:21
We want to prove the other direction.
02:23
So let's suppose that s and t are both final states or neither are final states.
02:27
This implies that f of s lambda and f t lambda are both final or are both non -final states, since lambda is the empty string, and this is true by definition.
03:19
Now, lambda is the only string of length 0, so it follows that f of s x and f of t x are both final or non -final states for all strings x of length less than are equal to zero.
04:07
Therefore, it follows that s is zero related to t.
04:12
This is what we wanted to prove.
04:16
So we've shown s is zero related to if only if t and s are both final states, and neither are final states.
04:34
Now we're asked to conclude that each final state of 1 over m, which is an r star equivalence class, contains only final states of m.
04:48
So let a be an equivalence class of m bar.
05:08
Now this implies that a is an r -star equivalence class of m.
05:31
So all states in a are either all final states or are all non -final states in m by a previous exercise.
06:08
And this implies that all final states of m bar are also final states in m.
06:43
Now in part b, we're asked to prove that if k is a positive integer, then it follows that s.
06:52
S is k related to t if and only if s is k minus 1 related to t and f of s a are k minus 1 f of t a for every a in the language.
07:23
So let k be a positive integer.
07:34
Now suppose that s is k related to t then it follows that f of s x and f of tx are both final states or both not final states.
07:52
For all strings x of length less than or equal to k.
08:22
Now let y be an input string of length k minus 1 and let a be an element of the input set.
08:49
So since y is an input string of length at most k, it follows that s is k minus 1 related to t.
09:07
Now, ya is the strength of length at most k.
09:19
Well, we have that f of s, a .y is equal to f of s...