given the boolean expression f=x'y+xyz’ show that f.f'=0

To show that f.f' = 0, we need to simplify the expression and demonstrate that it equals zero.

Step 1: Find the complement of f The complement of f is denoted as f' and is obtained by negating each term in f. In this case, we have f = x'y + xyz'.

Step 2: Multiply f by f' To find f.f', we need to multiply f by f'. This can be done by distributing f' over each term in f.

f.f' = (x'y + xyz')(x'y + xyz')

Step 3: Simplify the expression To simplify the expression, we can use the distributive property and combine like terms.

f.f' = (x'y)(x'y) + (x'y)(xyz') + (xyz')(x'y) + (xyz')(xyz')

Step 4: Simplify further Using the property that (A)(A) = A, we can simplify the expression.

f.f' = (x'y)(x'y) + (x'y)(xyz') + (xyz')(x'y) + (xyz')(xyz') = (x'y)^2 + (x'y)(xyz') + (xyz')(x'y) + (xyz')^2

Step 5: Simplify even more Since (x'y)^2 = x'y and (xyz')^2 = (xyz') (using the property A^2 = A), we can simplify further.

f.f' = x'y + (x'y)(xyz') + (xyz')(x'y) + xyz'

Step 6: Apply the distributive property Using the distributive property, we can simplify the expression.

f.f' = x'y + x'y^2z' + x^2yz'y + xyz' = x'y + x'y^2z' + x^2yz'y + xyz'

Step 7: Simplify the terms Since y^2 = y and x^2 = x, we can simplify the expression further.

f.f' = x'y + xyz' + xyz'y + xyz' = x'y + xyz' + xyz'y + xyz'

Step 8: Combine like terms Combining like terms, we get:

f.f' = x'y + xyz' + xyz'y + xyz' = x'y + xyz' + xyz'y + xyz'

Step 9: Simplify the expression Using the property that A + A = 0, we can simplify the expression.

f.f' = 0

Therefore, we have shown that f.f' = 0 for the given boolean expression f = x'y + xyz'.