Proof of sets $A = ({n \in \mathbb{Z}} | (\exists k \in \mathbb{Z})(n=4k+1)) $, $B = ({n \in \mathbb{Z}} | (\exists k \in \mathbb{Z})(n=4j-7)) $

Question

$A = ({n \in \mathbb{Z}} | (\exists k \in \mathbb{Z})(n=4k+1)) $

$B = ({n \in \mathbb{Z}} | (\exists k \in \mathbb{Z})(n=4j-7)) $

Prove that $A = B$

My attempt:

Show that $A \subseteq B$

Let $x \in A$. By definition of $A$, $x = 4m+1$ for some $m \in \mathbb{Z}$, Letting $n = m-2$, checking by substition that $4n-7 = 4(m-2) +1 = 4m-7.$ Therefore, $x = 4n-7$ for some $\mathbb{Z}$ Thus $x \in B$ and we have show $A \subseteq B$

Let $x \in B$. By definition of $B$, $x = 4m-7$ for some $m \in \mathbb{Z}$, Letting $n = m+2$, checking by substition that $4n-7 = 4(m+2) -7 = 4m+1.$ Therefore, $x = 4n+1$ for some $\mathbb{Z}$ Thus $x \in B$ and we have show $B \subseteq A$

Therefore $A = B$

Is this correct approach?

Answer 1

Yes, it is correct. This is a standard approach to showing that some sets are the same.