# How does the subtraction gate work in Fourier space

I am currently reading Shor's algorithm on my own and I come across a paper via this link. It shows the circuit for implementing Shor's alogrithm. Here it depicts that taking a QFT circuit on the number a on the qubits b that are already in fourier space would result in the addition of a and b in Fourier space.

The addition is quite intuitive. However, when it comes the subtraction, I am stuck.Why taking an inverse QFT circuit would result in the $$b-a$$ or $$2^{n+1} - (a-b)$$? Wouldn't it return $$a+b$$ as the second stage output is $$\Phi(a+b)$$?

In the paper they mention that a thick bar drawn on the left of a gate is their notation for its inverse. So in the second circuit they have drawn the inverse of $$\phi ADD(a)$$, i.e. $$\phi ADD(-a)$$.
$$QFT^{-1}$$ is there to un-Fourier transform the result, not to actually perform the substraction.