Verified answer

I'll try to explain the processes for the first in greater depth.

1:

p = A'B'C + A'BC + AB'C' + ABC'

Group first two terms and factor A' from each, and

group last two terms and factor A from each.

p = A'(B'C + BC) + A(B'C' + BC')

Factor C from parenthesized terms of first expression, and

factor C' from parenthesized terms of second expression.

p = A'(B' + B)C + A(B' + B)C'

Evaluate expression that are equivalent to the universe.

p = A'(1)C + A(1)C'

Eliminate universe terms.

p = A'C + AC'

Using a slightly different form of notation than what you've provided, the correctness of the simplifications can be checked using the Truth Table Generator at:

http://turner.faculty.swau.edu/mathematics/materia...

A'B'C + A'BC + AB'C' + ABC' = A'C + AC'

can be rewritten as

~A&~B&C + ~A&B&C + A&~B&~C + A&B&~C = ~A&C + A&~C

for input to the generator.

2:

p = A'B'C' + A'BC' + A'BC' + AB'C

= A'(B'C' + BC' + BC') + AB'C

= A'(B' + B + B)C' + AB'C

= A'(B' + B)C' + AB'C

= A'(1)C' + AB'C

= A'C' + AB'C

~A&~B&~C + ~A&B&~C + ~A&B&~C + A&~B&C = ~A&~C + A&~B&C

3:

p = A'BC'D + A'BCD + ABC'D + ABCD

= A'B(C' + C)D + AB(C' + C)D

= A'B(1)D + AB(1)D

= A'BD + ABD

= (A' + A)BD

= (1)BD

= BD

~A&B&~C&D + ~A&B&C&D + A&B&~C&D + A&B&C&D = B&D

4:

p = ABC'D + AB'C'D + A'BCD + AB'CD + A'BCD'

= ABC'D + AB'C'D + AB'CD + A'BCD + A'BCD'

= A(BC'D + B'C'D + B'CD) + A'(BCD + BCD')

= A(BC' + B'C' + B'C)D + A'B(CD + CD')

= A(BC' + B'[C' + C])D + A'BC(D + D')

= A(BC' + B'[1])D + A'BC(1)

= A(BC' + B')D + A'BC

= ABC'D + AB'D + A'BC

A&B&~C&D + A&~B&~C&D + ~A&B&C&D + A&~B&C&D + ~A&B&C&~D = A&B&~C&D + A&~B&D + ~A&B&C

(had to remove the blanks to check the last one...too long for input field)