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Find how many min NAND gates are required to realize the function :

1-  f(A,B,C)=A+BC

2-f(A,B,C,D)=(A'+B')(C+D)

please help me with the intuition required?

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We try to realize the expression using only AND and NOT by applying Demorgan's Law

1-  f(A,B,C)=A+BC

  ((A)' (BC)')'

So number of NAND Gates  = 1[for A'] + 1 [for (BC)'] + 1[for (A' (BC)')'] = 3 NAND gates are used

2-f(A,B,C,D)=(A'+B')(C+D)

                   = (AB)'.(C+D)

                   = (AB)'C + (AB)'D

                   = ( ((AB)'.C ) ' . ((AB)'D)' )'

 

= So number of NAND Gates = 1 [for (AB)'] + 1 [for ((AB)'.C ) '] + 1 [for (AB)'D)'] + 1 [for ( ((AB)'.C ) ' . ((AB)'D)' )'] = 4 NAND gates

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