Let A = { y € Z | y = 10b - 3 for some integer b } B = { z € Z | z = 10c + 7 for some integer c } Prove or Disprove that A = B

Let $A = \{ y\in\Z\ |\ y = 10b - 3\text{ for some integer }b\}$,

$B = \{ z \in \Z\ |\ z = 10c + 7\text{ for some integer }c \}$

Let us prove that $A = B.$

Let $x\in A.$ Then $x=10b-3$ for some integer $b.$ It follows that $x=10(b-1)+7,$ and $b-1$ is also integer. Therefore, $x\in B.$ Consequently, $A\subset B.$

Let $x\in B.$ Then $x=10c+7$ for some integer $c.$ It follows that $x=10(c+1)-3,$ and $c+1$ is also integer. Thus $x\in A.$ Therefore, $B\subset A.$

We conclude that $A=B.$