Add a small missed optimization: turn X == C ? X : Y into X == C ? C : Y. This

Add a small missed optimization: turn X == C ? X : Y into X == C ? C : Y. This
removes one use of X which helps it pass the many hasOneUse() checks.

In my analysis, this turns up very often where X = A >>exact B and that can't be
simplified unless X has one use (except by increasing the lifetime of A which is
generally a performance loss).

git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@128373 91177308-0d34-0410-b5e6-96231b3b80d8