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 :)