A subtract that keeps its order, mid-chain
The integer subf flips its operands on you. fsubs doesn't. It just computes fD = fA - fB, left to right, no surprises. That plain ordering matters most mid-chain, where you have to stay clear on which value is being subtracted from which.
Say delta(p, q, r) adds the first two arguments, then takes the third away:
fadds f0, f1, f2 # f0 = p + q
fsubs f1, f0, f3 # f1 = f0 - r = (p + q) - r
blr
The running total p + q sits in f0. fsubs leaves it on the left, so the result is (p + q) - r, not r - (p + q). Swap the instruction to fsubs f1, f3, f0 and the math swaps with it. Whatever order you read in the operands is the order you write in the C.
Same shape in the target, an add feeding a subtract. Read the operand order on the fsubs and you will know which value gets subtracted from which.
Your task
Write net3, taking three f32s, to reproduce the assembly above.