00:01
Okay, so here we have the statement, r is reflective and transitive, then this implies that rn is equal to r.
00:06
So what i'll do here is essentially a reductionist version of an induction proof, and we'll show that r squared is equal to r.
00:16
So really, the reason why you only really need to show this is because, let's suppose we had r cubed.
00:22
Well, r cubed, we can decompose that into being r composed with r squared.
00:27
R squared we're already known to be r, so this becomes r, composed with r which is r squared and this is just r and you can do this continually on from n.
00:36
So essentially to prove this is just to prove a very incredibly reductionist induction argument.
00:44
But nonetheless, so we let let um a be an arbitrary element in r.
00:53
So since r is reflexive, it must mean that b and b is in r as well because b must be in reset a and a switch a with b...