Their Explanation is given below from "Introduction to knot theory" of Richard H. Crowell.
And this was an example on them:
But I do not understand in the example, in the fifth line, from where we get this term $z(y^{-1} a)^{-1}$?
Also I do not understand How the last line of the example is reached from the line just before it, more precisely, why this term $z(y^{-1} a)^{-1}$ is removed?
Could anyone explain these questions for me please?
The word $z(y^{-1}a)^{-1}$ can be written as $y^{-1}(a(yz)^{-1})^{-1}y$, the conjugation of the inverse of $a(yz)^{-1}$ by $y^{-1}$. Recall that a consequence to a set of relations is any product of arbitrary conjugates of them (since the corresponding presented group is the quotient of a free group by the normal closure of the relations). Thus, move I applies.
In the penultimate line, there is the relation $z(y^{-1}a)^{-1}$. This means that in the quotient group, $[z]=[y^{-1}a]$ (where I'm just using brackets to distinguish between the word $y^{-1}a$ in the free group and $[y^{-1}a]$ as its image in the presented group). Since this implies $z$ is determined by $y^{-1}$ and $a$, it is safe to replace all occurrences of $z$ in the presentation by $y^{-1}a$ and delete that generator, which is move II'.