What is output of Y in the given circuit? I. BC+AC' II. (A+C)(B+C')

Both statements are correct.

The circuit simplifies to:

Y=(A+BC)(B+C′)

Expanding:

Y=AB+AC′+BBC+BCC′

Since BB=B and CC′=0,

Y=AB+AC′+BC

Now notice that AB is redundant:

AB=AB(C+C′)=ABC+ABC′

• ABC is already covered by BC

• ABC′ is already covered by AC′

Therefore,

Y=BC+AC′

Y=(A+BC)(B+C′)=AB+AC′+BC=BC+AC′

Also, factoring gives:

BC+AC′=(A+C)(B+C′)