Tietze Transformations: $\langle a, b, c\mid b^2, (bc)^2\rangle$ and $\langle x, y, z\mid y^2, z^2\rangle$. This is Exercise 1.5.2 of "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations," by W. Magnus et al.
The Details:
A Tietze transformation of one presentation $$\langle a, b, c\mid P, Q, R, \dots\rangle\tag{1}$$ of a group $G$ into another presentation of $G$ is one of the following:
(T1) Adding relators. If the words $S, T, \dots$ are derivable from $P, Q, R, \dots$ then add $S, T, \dots$ to the defining relators in $(1)$.
(T2) Removing relators. If some of the relators, say, $S, T, \dots$, listed among the defining relators $P, Q, R, \dots$ are derivable from the others, delete $S, T, \dots$ from the defining relators in $(1)$.
(T3) Adding generators. If $K, M, \dots$ are any words in $a, b, c, \dots$, then adjoin the symbols $x, y, \dots$ to the generating symbols in $(1)$ and adjoin the relations $x=K, y=M, \dots$ to the defining relators in $(1)$.
(T4) Removing generators. If some of the defining relations in $(1)$ take the form $p=V, q=W, \dots$, where $p, q, \dots$ are generators in $(1)$ and $V, W, \dots$ are words in the generators other than $p, q, \dots$, then delete $p, q, \dots$ from the generators, delete $p=V, q=W, \dots$ from the defining relations, and replace $p, q, \dots$ by $V, W, \dots,$ respectively, in the remaining defining relators in $(1)$.

Theorem 1: If a presentation can be transformed into another by Tietze transformations, then the two are isomorphic.

The Question:

Show by means of Tietze transformations that the presentations $$\langle a, b, c\mid b^2, (bc)^2\rangle\tag{A}$$ and $$\langle x, y, z\mid y^2, z^2\rangle\tag{B}$$ define isomorphic groups.

My Attempt:
This is an application of Theorem 1.
By (T3), $(A)$ is $$\langle a, b, c, x\mid b^2, (bc)^2, a=x\rangle,$$ which is $$\langle x, b, c\mid b^2, (bc)^2\rangle$$ by (T4).
Now what?
I guess I could let $y=b$ and $z=bc$ somehow - by (T3)? - but I don't know how to do this explicitly, using (T1) - (T4) to get $(B)$ from $(A)$.
Please help :)
 A: T3 to add generator $y$ and relation $y = b$: $\langle x,b,c,y | b^2, (bc)^2, y^{-1}b \rangle$
T1 to add relation $(yc)^2 = 1$ (derivable from $y = b$ and $(bc)^2 = 1$): $\langle x,b,c,y | b^2, (bc)^2, y^{-1}b, (yc)^2 \rangle$
T2 to remove relator $(bc)^2$: $\langle x,b,c, y | b^2, y^{-1}b, (yc)^2 \rangle$
T1 to add relator $y^2$: $\langle x,b,c,y | b^2, y^{-1}b, (yc)^2, y^2 \rangle$
T2 to remove relator $b^2$: $\langle x,b,c,y | y^{-1}b, (yc)^2, y^2 \rangle$
Now there is only one relator involving $b$ and it has the form $bw$ for a word $w$ not involving $b$, then, we may remove it.
T4 to remove generator $b$: $\langle x,c,y | (yc)^2, y^2 \rangle$
T3 to add generator $z$ with relation $z = yc$: $\langle x,c,y, z | (yc)^2, y^2, z^{-1}yc \rangle$
Now, $z = yc$ and $(yc)^2 = 1$ imply $z^2 = 1$, so, we may add this relation:
T1 to add relator $z^2$: $\langle x,c,y, z | (yc)^2, y^2, z^{-1}yc, z^2 \rangle$
Now, $(yc)^2 = 1$ is derivable from $z = yc$ and and $z^2 = 1$, so, we may remove it:
T2 to remove relator $(yc)^2$: $\langle x,c,y, z | y^2, z^{-1}yc, z^2 \rangle$
T4 to remove generator $c$: $\langle x,y, z | y^2, z^2 \rangle$
