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BSD 4.3reno

The pattern looks like this:

	movl	P,rA
	xxxl3	Q,rA,rB
	yyyl2	R,rB

xxx and yyy may be any of several binary operators like 'add',
'mul', 'and'.  rA and rB must both be temporary registers.

Here's how it happens...  We work backwards from the last instruction:

	movl	P,rA
	xxxl3	Q,rA,rB
=>	yyyl2	R,rB

Here we note that rB is a 'modify' operand and mark rB alive.

	movl	P,rA
=>	xxxl3	Q,rA,rB
	yyyl2	R,rB

We see that this instruction writes into rB and kills it, but since
rB is previously alive, the store is not redundant and we leave it
alone.

=>	movl	P,rA
	xxxl3	Q,rA,rB
	yyyl2	R,rB

We notice that this is a useless store into rA, since we can replace
rA with P in the following instruction.  We delete this instruction
and modify the following one.

=>	xxxl3	Q,P,rB
	yyyl2	R,rB

Unfortunately rB is no longer alive because we killed it evaluating
the earlier form of this very instruction.  The optimizer assumes that
we have a redundant store and deletes the instruction.  Bleah.

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