Here, first we will prove (i) ⬄ (ii)

Where, (i) = A ⊂ B and (ii) = A – B ≠ ϕ

Let us assume that A ⊂ B

Now, we need to prove A – B ≠ ϕ

If possible, let A – B ≠ ϕ

Thus, there exists X ϵ A, X ≠ B, but this is impossible as A⊂ B

∴ A – B = ϕ

And A⊂ B =>A – B ≠ ϕ

Let us assume that A – B ≠ ϕ

Now, to prove: A ⊂ B

Let Xϵ A

It can be concluded that X ϵ B (if X ∉ B, then A – B ≠ ϕ)

Thus, A – B = ϕ => A ⊂ B

∴(i) ⬄ (ii)

Let us assume that A ⊂ B

To prove: A ∪ B = B

⇒ B ⊂ A ∪ B

Let us assume that, x ϵ A∪ B

⇒ X ϵ A or X ϵ B

Taking Case I: X ϵ B

A ∪ B = B

Taking Case II: X ϵ A

⇒ X ϵ B (A ⊂ B)

⇒ A ∪ B ⊂ B

Let A ∪ B = B

Let us assume that X ϵ A

⇒ X ϵ A ∪ B (A ⊂ A ∪ B)

⇒ X ϵ B (A ∪ B = B)

∴A⊂ B

Thus, (i) ⬄ (iii)

Now, to prove (i) ⬄ (iv)

Let us assume that A ⊂ B

It can be observed that A ∩ B ⊂ A

Let X ϵ A

To show: X ϵ A∩ B

Since, A ⊂ B and X ϵ B

Thus, X ϵ A∩ B

⇒ A ⊂ A ∩ B

⇒ A = A ∩ B

Similarly, let us assume that A ∩ B = A

Let X ϵ A

⇒ X ϵ A ∩ B

⇒ X ϵ B and X ϵ A

⇒ A ⊂ B

∴ (i) ⬄ (iv)

Hence, proved that (i)⬄ (ii)⬄ (iii)⬄ (iv)