How the expression !A + (A . !B) = !(A.B)?

Question

I have an expression !A+(A.!B) and on an expression solver, it gives the !A+(A.!B) = !(A.B)?. The solver notified that "Apply the Absorption Law" A.B+!A = B+!A.

I have made the truth tables for both the expressions and the answer was correct. But the problem is I can not understand how the absorption law has got implemented to my expression !A+(A.!B)?

Can someone please explain in details how the absorption law has got implemented to my expression?

janw · Accepted Answer · 2020-04-21T09:55:36.057

1

I am going to assume that + = OR, . = AND and ! = NOT.

The absorption law was applied in the very first step:

  !A + A.!B
= !A + !B     (if the first monomial does not hold, A is "true"
               and thus does not need to be checked again)
= !(A.B)      (De Morgan's rule)

edited Apr 21 '20 at 09:55

answered Apr 21 '20 at 09:34

janw

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Can you please explain how the absorption law gets used on my expression? – Khuram Apr 21 '20 at 09:52
Actually I have understood that but saying for someone else's reference – Khuram Apr 21 '20 at 09:54
@MKJ I updated the answer to directly mention the absorption law. – janw Apr 21 '20 at 09:56

How the expression !A + (A . !B) = !(A.B)?

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