A minor issue with register allocator

[sig]

Ohh, another issue.

Having an input intermediate:
#(#block 1 #(#push #ebp) #(#mov #esp #ebp))
#(#block 3 #(#mov #(#mem #(#add #ebp -4)) 't1') #(#mov #(#mem #(#add
#ebp -8)) 't2'))
#(#block 5 #(#mov #(#add 't1' 't2') 't3') #(#jmp #block4))
#(#block 4)
#(#block 6 #(#mov 't3' #eax) #(#jmp #block2))
#(#block 2 #(#mov #ebp #esp) #(#pop #ebp) #(#ret))

after running LowLevelOptimizer and InstructionSelector i got:

#(#block 1 #(#push #ebp) #(#mov #esp #ebp))
#(#block 3 #(#mov #(-4 #ebp) 't1') #(#mov #(-8 #ebp) 't2'))
#(#block 5 #(#mov 't2' 't4') #(#add 't1' 't4') #(#mov 't4' 't3')
#(#jmp #block4))
#(#block 4)
#(#block 6 #(#mov 't3' #eax) #(#jmp #block2))
#(#block 2 #(#mov #ebp #esp) #(#pop #ebp) #(#ret))

and then after coloring register allocator:
#(#(#block 1 #(#push #ebp) #(#mov #esp #ebp))
#(#block 3 #(#mov #(-4 #ebp) #eax) #(#mov #(-8 #ebp) #eax))
#(#block 5 #(#add #ebx #eax) #(#jmp #block4))
#(#block 4)
#(#block 6 #(#jmp #block2))
#(#block 2 #(#mov #ebp #esp) #(#pop #ebp) #(#ret)))

take a look at block3 , as you see it chooses #eax for replacing
temporary registers in both instructions (which is wrong of course)
and then in block5 , it seems thinks that there was #ebx used in one of them.