Prove both the DeMorgan’s laws using truth tables?

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1 Answer
Mar 12, 2018

Answer:

See below

Explanation:

Let #x in(AuuB)^c# then #x!inAuuB#. That means #x!inA and x!inB#. This is the same to say #x inA^c and x inB^c#. We can write both conditions in that way: #x in A^cnnB^c#

Thus, the inclusion #(AuuB)^csubA^cnnB^c# (1) is true.

Lets see the oposite. That will prove #(AuuB)^c=A^cnnB^c#

Let #x in A^cnnB^c#. That means #x in A^c and x in B^c#. This it's the same to say #x!inA and x !inB#. Then #x!inAuuB#. And finally, that it's the same to say #x in(AuuB)^c#

We have prove #(AuuB)^csupA^cnnB^c# (2)

Because (1) and (2) are true simultaneusly, then #(AuuB)^c=A^cnnB^c# QED