How to simplify this boolean algebra (logic)

[aruwin]

This is related to logic circuits.
How do I simplify this boolean algebra(logic) into its simplest expression?

h = (abc + abd + acd + bcd)’

My first attempt would be to change it into product-of-sum.
h = (abc)'(abd)'(acd)'(bcd)'

and then what next?

 

[i_madini]

This is related to logic circuits.
How do I simplify this boolean algebra(logic) into its simplest expression?

h = (abc + abd + acd + bcd)’

My first attempt would be to change it into product-of-sum.
h = (abc)'(abd)'(acd)'(bcd)'

and then what next?

h = (abc)'(abd)'(acd)'(bcd)'

= (a'+b'+c')(a'+b'+d')(a'+c'+d')(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'+c'+d')(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'+c'+d')


I hope it is correct..

 

[aruwin]

h = (abc)'(abd)'(acd)'(bcd)'

= (a'+b'+c')(a'+b'+d')(a'+c'+d')(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'+c'+d')(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'+c'+d')


I hope it is correct..

Could you explain to me how you add the new terms into the expression?

 

[i_madini]

Let's assume we have :

F = A'+AB

= A'(B+B') + AB # B+B'= 1, will not affect the equation.

= A'B + (A'B' + AB) # Cancel each other.

= A'B

By the same way :)

 

[rezeew]

The next step would be to use De Morgan's laws to simplify the expression further. We can apply De Morgan's laws to the parentheses, which would result in the following expression:

h = (a'+b'+c')(a'+b'+d')(a'+c'+d')(b'+c'+d')

We can then use the distributive property to expand the expression and eliminate any redundant terms. This would result in the simplest expression for h:

h = a'b'c' + a'b'd' + a'c'd' + b'c'd'