NMR magnetically equivalent protons for a 1,4-disubstituted benzene ring

Below I have drawn a molecule with random substituents $$\ce{X}$$ and $$\ce{Y}$$:

If the distances from $$\ce{H_b}$$ and $$\ce{H_a}$$ to $$\ce{H_d}$$ are considered, they are different - $$\ce{H_b}$$ is 5 bonds away and $$\ce{H_a}$$ is 3 bonds away, so $$\ce{H_b}$$ sees $$\ce{H_d}$$ differently compared to $$\ce{H_c}$$. The same argument can be made vice versa and also between the same pair of protons towards $$\ce{H_c}$$. This was what the lecturer told me regarding this compound below.

This led me to believe then that all the protons in the molecule are magnetically inequivalent and using up to $$\ce{^4J}$$ coupling, all protons show a doublet of doublet.

However, when I was in school, I was told that if symmetry was a property of a molecule, as above, $$\ce{H_a}$$ and $$\ce{H_b}$$ were equivalent, as are $$\ce{H_c}$$ and $$\ce{H_d}$$, and each pair of protons gives a doublet. Researching this matter on Google still held true in my undergraduate studies.

What am I getting wrong here?

I read a quote on another question highlighted here

For two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other non-chemical shift equivalent nuclei in the molecule.

So as $$\ce{H_a}$$ and $$\ce{H_b}$$ are chemically equivalent but $$\ce{H_b}$$ couples differently to $$\ce{H_c}$$ than $$\ce{H_a}$$ does, that therefore means that surely $$\ce{H_a}$$ and $$\ce{H_b}$$ are magnetically inequivalent?

It's not as simple as a dd either; when you have magnetically inequivalent protons that see each other, the first-order rules (the so-called $$n+1$$ rule) doesn't work so well anymore. You can see some real-life examples of how these molecules look like here: https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-15-AABB.htm For the p-disubstituted benzene it sort of looks like a doublet if you zoom out far enough, but if you look closely, there are additional peaks flanking the major peaks.
• Ah right that makes sense then. So if I considered $\ce{^4J}$ coupling as I mentioned under first-order rules (because this is being examined soon) I would be correct given the information I know of at this moment in time in my studies? – vik1245 Jan 1 '20 at 2:41