I agree on the input diodes but
not on why the extra inverters (or output stages).
Look at picture 3 at the bottom. I showed how to remove 4 transistors and get the same function.
A transistor pair makes a inverter. Two pair make a buffer. Some times the schematic does not show the "buffer" because it has no logic function. It does have a function of increasing the gain!.
There are two types of parts like this. CD4001ub Unbuffered and D4001b Buffered.
Your circuit is buffered. By removing the 4 transistors you get unbuffered.
The graph on the right is buffered. There are 3 lines for 5V, 10V and 15V supply. Lets look at the 10V supply. The output voltage swings from 0 to 10 volts as the input changes. You can see at 5V on the input the output switches. Below 5=0 and above 5=1. There is a very small area from 5 to 5.2 where the part is not certain if it is a 1 or a 0.
Now look at the left graph. Going from 0 to 1 the part switches at about 7.0 volts. BUT Going from 1 to 0 the part switches at 3.0 volts.
Also because of the lower gain the part has a much wider uncertainty area. Going from 0 to 1 at Vin=2.5V the output is not at one of the supplies but is about 2 volts off from the supply. (This part could leave its output at 2V or 1V or something not 0V. )
There is a place to use UB and a place to use B.