Step 1

Given: \(\displaystyle{A}={\left\lbrace{x}\in{Z}:\exists_{{{y}\in{z}}}{x}={4}{y}+{1}\right\rbrace}\)

\(\displaystyle{B}={\left\lbrace{x}\in{Z}:\exists_{{{y}\in{z}}}{x}={8}{y}-{7}\right\rbrace}\)

To prove or disprove: \(\displaystyle{A}\subseteq{B},{B}\subseteq{A}\)

Step 2

Counter Example:

Reason: Let \(\displaystyle{5}\in{A}\)

\(\displaystyle{5}={4}{\left({1}\right)}+{1}\)

But \(\displaystyle{5}={8}{y}-{7}\)

\(\displaystyle\Rightarrow{\frac{{{3}}}{{{2}}}}={y}\)

\(\Rightarrow \frac{3}{2} \not{\in} Z\)

\(\Rightarrow 5 \not{\in} B\)

Therefore, \(A \not{\subseteq} B\)

Step 3

Counter Example:

Reason: Let \(\displaystyle{40}\in{B}\)

\(\displaystyle{40}={8}{\left({5}\right)}-{7}\)

But \(\displaystyle{40}={4}{y}+{1}\)

\(\Rightarrow \frac{39}{4}=y\)

\(\Rightarrow \frac{39}{4} \not{\in} Z\)

\(\Rightarrow \not{\in} A\)

Therefore, \(B \not{\subseteq} A\)