= Solution
The <row rank> of $A$ is the dimension of the span of its rows in $\mathbb R^n$; its <column rank> is the dimension of the span of its columns in $\mathbb R^m$. <Elementary row operations> replace rows by invertible <linear combinations>, so preserve their span. They also preserve every column relation: $Ac=0$ if and only if $EAc=0$ for an invertible <elementary matrix> $E$. Reduce $A$ to <row echelon form> $R$, with $r$ nonzero rows. These rows are independent because their leading entries occur in distinct, successively later columns. The $r$ pivot columns are independent by their triangular pivot submatrix, while every column lies in the $r$-dimensional coordinate space supported on the nonzero rows. Thus both ranks of $R$, and hence both ranks of $A$, equal $r$. \b[<Row rank> equals <column rank>.]
<Elementary column operations> replace columns by invertible <linear combinations> and preserve their span. Equivalently, right multiplication by an invertible <elementary matrix> $F$ preserves row relations, since $c^TAF=0$ if and only if $c^TA=0$. Therefore either <elementary row operations> or <elementary column operations> preserve the <matrix rank>, and any finite sequence gives $\boxed{\operatorname{rank}(A')=\operatorname{rank}(A)}$.
For the block <matrices>, put $M=\begin{pmatrix}PQ&0\\Q&QR\end{pmatrix}$. The explicit invertible multiplications
$$
\begin{pmatrix}I&-P\\0&I\end{pmatrix}M\begin{pmatrix}I&-R\\0&I\end{pmatrix}=\begin{pmatrix}0&-PQR\\Q&0\end{pmatrix}
$$
follow by first subtracting $P$ times the lower block row from the upper one and then subtracting the first block column times $R$ from the second. Changing the sign of the upper block row gives $N=\begin{pmatrix}0&PQR\\Q&0\end{pmatrix}$. These operations preserve <matrix rank>, proving that the two specified block <matrices> have equal rank. Their off-diagonal form gives $\operatorname{rank}(N)=\operatorname{rank}(Q)+\operatorname{rank}(PQR)$, since the two image components occupy independent coordinate blocks.
Choose $r_1=\operatorname{rank}(PQ)$ independent columns of $PQ$ and $r_2=\operatorname{rank}(QR)$ independent columns of $QR$. Select the corresponding $r_1$ columns from the first block column of $M$ and $r_2$ from the second. In a <linear relation> among these selected columns, the upper coordinate block forces every coefficient of the first selection to vanish; the lower block then forces every coefficient of the second to vanish. Consequently $\operatorname{rank}(M)\geq r_1+r_2$. Combining this with the equality of block ranks proves the <Frobenius rank inequality>:
$$
\boxed{\operatorname{rank}(PQ)+\operatorname{rank}(QR)\leq\operatorname{rank}(Q)+\operatorname{rank}(PQR).}
$$
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