Why is the substrate terminal considered to have no effect on MOSFET operation?

[sandy.bridge]

I will quote my textbook, and further I will explain what I do not understand. "Since, as we shall see shortly, the drain will always be at a positive potential relative to the source, the two pn junctions can effectively be cut off by connecting the substrate terminal to the source terminal. Thus, here, the substrate will be considered as having no effect on device operation and we will further consider MOSFET's to be three terminal devices..."
My textbook makes this assumption and has no explanation as to why it makes this assumption, and it also neglects explaining how connecting the source and the substrate terminals will effectively cut off the pn junctions. I understand how pn junctions work, but this is not clicking.

 

[Born2bwire]

The substrate and source/drain forms a p-n junction. So there is the potential of having a voltage between the source/drain and the substrate (or body) that allows for the forward bias mode to exist and current to flow. However, if we short the body to the source pin, then the voltage across the source-body p-n junction is zero and we just have the typical equilibrium case. Given that the drain is always at a positive potential relative to the source, then the drain is also at a positive potential relative to the body and thus the drain-body p-n junction is reverse biased.

Finally, when we apply a voltage across the gate and the body to create an inversion layer between the drain and source, the inversion layer is the same doping as the drain and source. So in this manner we can prevent any current from flowing between the source/drain and the body regardless of the current transistor mode.