Let s be a given set and a \subset s and b \subset s. prove that a \subset b \leftrightarrow s\a \supset s\b.

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...

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.