Question

a'b'c'+abc'+a'bc+a'b'c+abc=a'b'+b(a+c)​

Answer 1

Answer:

A′B′C+A′BC′+AB′C′+ABC=A′B′C′+ABC+A′BC′+AB′C′  

C(A′B′+AB)+C′(A′B+AB′)

C((A′B′+AB)′)′+C′(A′B+AB′)

Using Demorgan’s Law

C((A+B)(A′+B′))′+C′(A⊕B)

C(AA′+AB′+BA′+BB′)+C′(A⊕B)

C(0+AB′+A′B+0)′+C′(A⊕B)

C(AB′+A′B)+C′(A⊕B)

C(A⊕B)′+C′(A⊕B)

Now let assume (A⊕B)=X

CX′+C′X

Which is C⊕X

Or X⊕C

Now substituting back X=(A⊕B), we get

(A⊕B⊕C)

which is required simplification

Where ⊕ denotes XOR

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Vijay Mankar (विजय मानकर), Teaching (1994-present)

Answered December 28, 2018

A′B′C+A′BC′+AB′C′+ABC

=A′B′C+ABC+A′BC′+AB′C′

=(A′B′+AB)C+(A′B+AB′)C′

=(A⊕B¯¯¯¯¯¯¯¯¯¯¯¯¯¯)C+(A⊕B)C′

=A⊕B⊕C

Answered March 17, 2019

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