The given function f(x,y, z) = (x + y) + (yz+ y’z’ +y’z)
IF we assume (x + y) is Part 1
and (yz+ y’z’ +y’z) is Part 2
- Here for (x+y) there 1 OR gate is required.
- for yz there is 1 AND gate and for y'z there will be another AND gate, so total 2 AND gate is required.
- and we know y'z'=(y+z)' which is nothing but a NOR gate.
- now to add yz, y'z' and y'z 1 OR gate required.
- And lastly to add part 1 and part 2 another OR gate required.
So in total 3 OR gate (b=3), 2 AND gate (a=2) and 1 NOR gate (c=1) is Required.
So (a+b)-c = 2+3-1 = 4
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But if we minimized the given expression f(x,y, z) = (x + y) + (yz+ y’z’ +y’z)
= x+y+yz+y’z’ +y’z
= x+y(1+z)+y’(z’+z)
=x+y+y’ (∵ 1+ anything = 1 and a+a’=1)
= x+1
=1
So we can clearly see if we minimize the given expression then No gates required.
then (a+b)-c=0