CODEWITHSUNDEEP
booleanboolean-expression
Dilan Madushanka
Dilan Madushanka

Reputation: 31

Simplify the Boolean Expressions (x+y).(x+z)

Simplify the Boolean expression " (x+y).(x+z) " .

I think answer is " x+y.z " But i don't know how t get that.

Upvotes: 1

Views: 33216

Answers (4)

Dhruv Pal
Dhruv Pal

Reputation: 959

(x+y)(x+z) 
= xx + xz + yx + yz 
= x + xz + yx + yz (since xx = x eg 0.0 = 0 , 1.1 = 1)
= x(1 + z + y) + yz
= x(1 + y) +yz (since 1 + z = 1 e.g 1+0 = 1 or 1+1 = 1)
= x(1) + yz (since 1 +y =1 as explained above)
= x + yz

Upvotes: 3

Alexander Crescent
Alexander Crescent

Reputation: 570

Here is a much simpler solution by using idempotent(xx = x) and absorption(x+xy = x) laws.

(x+y)(x+z) = xx+xz+xy+yz = x+yz

Upvotes: -1

ThomasPM
ThomasPM

Reputation: 61

(x+y)(x+z) -Distribute-> xx+xy+xz+yz -x.x=x-> x+xy+xz+yz -> x+x(y+z)+yz -x=x.1-> x.1+x(y+z)+yz -> x(1+(y+z))+yz -1+(y+z)=1-> x+yz

Upvotes: 2

Felipe Toshio Hassunuma Saito
Felipe Toshio Hassunuma Saito

Reputation: 46

You should use the De Morgan Law (A+B)=(A'.B'). It works this way: (X+Y)=X'.Y' and (X+Z)=X'.Z' By commutativity: (X+Y).(X+Z)=(X'.Y').(X'.Z')=X'.Y'.X'.Z'=X'.X'.Y'.Z' By idempotence: X'.X'=X' Then: X'.X'.Y'.Z'=X'.Y'.Z'=X'.(Y'.Z') Calling: Y'.Z'=W Then: X'.(Y'.Z')=X'.W' By De Morgan: X'.W'=(X+W) (I) Negating the affirmation: W'=Y'.Z' then W=(Y'.Z')'=Y'+Z'=Y.Z (II) By (I) and (II): (X+Y).(X+Z)=X+(Y.Z)=X+Y.Z

Upvotes: 3

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