Answer:
Here,
S is a set so that A ⊂ S and B ⊂ S,
We have to prove that,
A ⊂ B ⇔ S\A ⊂ S\B.
Suppose,
A ⊂ B
Also, let x ∈ S\B, where, x is an arbitrary,
⇒ x ∈ S but x ∉ B
⇒ x ∈ S but x ∉ A ( ∵ A ⊂ B )
⇒ x ∈ S\A
⇒ x ∈ S\B ⇒ x ∈ S\A
⇒ S\A ⊂ S\B.
Conversely,
Suppose, S\A ⊂ S\B,
Let y ∈ B, where y is an arbitrary,
⇒ y ∉ S - B
⇒ y ∉ S - A ( ∵ S\A ⊂ S\B )
⇒ y ∈ A but y ∉ S
⇒ y ∈ B ⇒ y ∈ A
⇒ A ⊂ B
Hence, proved...
