A longer running total
With four operands you have got three operations, hence three arithmetic instructions. A partial result builds up in r0, passed along from each instruction to the one after it, and the final instruction writes r3 to hand the answer back.
Here is delta(p, q, r, s). It runs two subtracts and then an add.
subf r0, r4, r3 # r0 = r3 - r4 = p - q
subf r0, r5, r0 # r0 = r0 - r5 = (p - q) - r
add r3, r6, r0 # r3 = s + ((p - q) - r)
blr
Every instruction grabs whatever r0 held and applies the next operation. The subf reversal is in force the whole way down, since subf rD, rA, rB is always rB − rA, which means the most recent r0 becomes the minuend of the next subf. So count the instructions and you know the number of operations. Decode them one by one and the chain reads off left-to-right.
This lesson's target leans on the same plan. Pin down which register carries which argument (r3→a, r4→b, r5→c, r6→d), then follow the accumulator instruction by instruction until the expression falls out.
Your task
Write addsub4, taking four ints, to reproduce the assembly above.