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De Morgan’s law

21 Nov

Hey so a bit of math here.

So I want to prove:  (A U B)’ = A’ ∩ B’ —> De Morgan’s law.

General case: In set theory, at least how I remember it is, if A = B then A is a subset of B and B a subset of A. Which is really obviously true. In mathy terms A ⊆ B and B ⊆ A. (⊆ means subset).

So what we want to do here is prove that (A U B)’ ⊆ A’ ∩ B’ and then prove also A’ ∩ B’ ⊆ (A U B)’ – just reversed the order.

STEP 1 – (A U B)’ ⊆ A’ ∩ B’

General case: The prove that A ⊆ B (A is a subset of B)  we show that there is a value common to both A and B.

suppose x ∈ (A U B)’ (∈ means ‘is contained in’). This wuld mean that x ∉ A U B and hence x ∉ A and x ∉ B (draw a Venn diagram if this is not too clear). which leads us to x ∈ A’ and x ∈ B’when combined is x ∈ A’ ∩ B’

And hence we have proven that x is in both A’ ∩ B’ and (A U B)’ which means (A U B)’ ⊆ A’ ∩ B’

STEP 2 – A’ ∩ B’ ⊆ (A U B)’

Try this on your own.

……………………

and finally (A U B)’ = A’ ∩ B’

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Posted by on November 21, 2012 in Uncategorized

 

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