I'm so confused about beq function in assembly MIPS.
Lets assume I have beq $t1 $t2 60 , so it would be mapped as I type function. so we could say for example assuming that beq mapped to  
100100 , $t1=5 , $t2=4, and immediate 60
so the block of IP type instruction is:
100100 | 5 | 4 | 60
but what's confused me that beq means if $t1 = $t2 then it would jump 60 steps from the offset PC ..
so the I type instruction should be mapped to 
100100 | 5 | 5 | 60
so the implementation mapping of registers should be 5 and 5 .. not 5 and 4 , am I right? because beq means $t1=$t2 ...so must the mapping instruction of registers of that instruction must be   
100100 | 5 | 5 | 60 and not
100100 | 5 | 4 | 60
so what should that instruction mapped?
to 100100 | 5 | 4 | 60
or to 100100 | 5 | 5 | 60 ?
 
     
    