Question

If A and B are the two non-empty sets , then {x ∶ x ∈A but x ∉B } ​

Answer 1

Hello Mate!!!!

Step-by-step explanation:

Let M = (A ∩ B)' and N = A' U B'

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ N

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'⇒ y ∈ A' or y ∈ B'

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'⇒ y ∈ A' or y ∈ B'⇒ y ∉ A or y ∉ B

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'⇒ y ∈ A' or y ∈ B'⇒ y ∉ A or y ∉ B⇒ y ∉ (A ∩ B)

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'⇒ y ∈ A' or y ∈ B'⇒ y ∉ A or y ∉ B⇒ y ∉ (A ∩ B)⇒ y ∈ (A ∩ B)'

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'⇒ y ∈ A' or y ∈ B'⇒ y ∉ A or y ∉ B⇒ y ∉ (A ∩ B)⇒ y ∈ (A ∩ B)'⇒ y ∈ M

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'⇒ y ∈ A' or y ∈ B'⇒ y ∉ A or y ∉ B⇒ y ∉ (A ∩ B)⇒ y ∈ (A ∩ B)'⇒ y ∈ MTherefore, N ⊂ M …………….. (ii)

Let M = (A ∩ B)' and N = A' U B'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A' or x ∈ B'⇒ x ∈ A' U B'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'⇒ y ∈ A' or y ∈ B'⇒ y ∉ A or y ∉ B⇒ y ∉ (A ∩ B)⇒ y ∈ (A ∩ B)'⇒ y ∈ MTherefore, N ⊂ M …………….. (ii)Now combine (i) and (ii) we get; M = N i.e. (A ∩ B)' = A' U B'

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Answer 2

Step-by-step explanation:

the set will be A-b... the condition is given...

Elements lies in set A not in B.

hope it helps..