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Prove that B \ (A \ C) ⊆ (B \ C) ∪ A ⇔ (B ∩ C) ⊆ A

We can rewrite B \ (A \ C) as B \ (A \ C) = B ∩ (A \ C)' = B ∩ (A ∩ C')' = (applying De Morgans Law) B ∩ (A' ∪ C) = (distributivity law) (B ∩ C) ∪ (B ∩ A')

Now we must prove that (B ∩ C) ∪ (B ∩ A') ⊆ (B \ C) ∪ A ⇔ (B ∩ C) ⊆ A.

(⇒) Let x ∈ (B ∩ C). Then clearly x ∈ (B ∩ C) ∪ (B ∩ A') and so from our premise it follows x ∈ (B \ C) ∪ A. In particular x ∈ (B \ C) or x ∈ A. But x ∉ (B \ C) as we said x ∈ (B ∩ C). So x ∈ A as required and so therefore if (B ∩ C) ∪ (B ∩ A') ⊆ (B \ C) ∪ A then (B ∩ C) ⊆ A.

(<=) (B ∩ C) ⊆ A ⇒ (B ∩ C) ∪ (B ∩ A') ⊆ (B \ C) ∪ A Case 1 : Let x ∈ (B ∩ C) then as (B ∩ C) ⊆ A it follows x ∈ A and so clearly it can be said x ∈ (B \ C) ∪ A as required.

Case 2: Let x ∈ (B ∩ A') then x ∈ B and x ∉ A. As x ∉ A from our premise x ∉ (B ∩ C). So x ∉ (B ∩ C) and x ∈ B meaning x ∈ (B\C) and so it can be said x ∈ (B\C) ∪ A as required. Therefore in either case we can conclude that (B ∩ C) ⊆ A ⇒ (B ∩ C) ∪ (B ∩ A') ⊆ (B \ C) ∪ A

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  • $\begingroup$ If it is the same as the one you posted two hours ago then it is correct. $\endgroup$ – drhab Nov 10 '14 at 20:10
  • $\begingroup$ LOL thank you and yes it is $\endgroup$ – Namch Nov 10 '14 at 20:10

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