Pgee 2017

Question

What is the equivalent de morgan's representation of the logic statement (c!=n && q <=5) ?

1. ! ( c!=n || q>=5)
2. ! ( c==n || q>=5)
3. ! ( c!=n || q>5)
4. ! ( c==n || q>5)

Answer 1

re writing it as ((c!=n)&&(q<=5))

now again rewriting it (((c!=n)&&(q<=5))')'

!((c!=n)&&(q<=5))'
!($\left ( \overline{c!=n} \right )||\left ( \overline{q<=5} \right )$)

!(c==n || q>5)

Answer 2

Demorgan's Theorem

¬(p ∧ q) ≡ ¬p ∨ ¬q
¬(p ∨ q) ≡ ¬p ∧ ¬q

As per Question:!(c!=n&&q<=5)

                            =!((c!=n)&&(q<=5))

                             =!((c==n)||(q>5))

                             =!(c==n||q>5)

Answer  4 is correct.