How Can Boolean Identities Simplify AB'CD+(ABC')'+ABCD'?

[Duderonimous]

Homework Statement

Minimize the following using boolean identities
1. AB'CD+(ABC')'+ABCD'

Homework Equations

Identity 1A=A 0+A = A
Null (or Dominance) Law 0A = 0 1+A = 1
Idempotence Law AA = A A+A = A
Inverse Law AA = 0 A+A = 1
Commutative Law AB = BA A+B = B+A
Associative Law (AB)C = A(BC) (A+B)+C = A+(B+C)
Distributive Law A+BC = (A+B)(A+C) A(B+C) = AB+AC
Absorption Law A(A+B) = A A+AB = A
DeMorgan's Law (AB) = A+B (A+B) = A B

The Attempt at a Solution

I'm going to use lower case letters now.f=ab'cd+(abc')'+abcd'
f'=(ab'cd+(abc')+abcd')'
=(ab'cd)'(abc')''(abcd')'
=(a'+b+c'+d')(abc')(a'+b'+c'+d)
=(aa'bc'+abbc'+abc'c'+abc'd')(a'+b'+c'+d)
=(abc'+abc'+abcd')(a'+b'+c'+d)
=(abc'+abcd')(a'+b'+c'+d')
=(0+0+0+0+abc'c'+abc'c'd'+abc'd+0)
=abc'+abc'd'+abc'd
=abc'+abc'(d'+d)
=abc'+abc'(1)
=abc'
f'=abc'
f=(abc')'

--> f=a'+b'+c

Do this look correct? If so is there a shorter way to minimize it? Is there a way to minimize without using DeMorgan's theorem at the top? Thanks

 

[maline]

I didn't read you calculation, but it seem clear that a and c must both be true, sp your final answer must be wrong. Try actually thinking about the logic, & afterward pick the formal identities to back up your intuition

 

[berkeman]

Homework Statement

Minimize the following using boolean identities
1. AB'CD+(ABC')'+ABCD'

Homework Equations

Identity 1A=A 0+A = A
Null (or Dominance) Law 0A = 0 1+A = 1
Idempotence Law AA = A A+A = A
Inverse Law AA = 0 A+A = 1
Commutative Law AB = BA A+B = B+A
Associative Law (AB)C = A(BC) (A+B)+C = A+(B+C)
Distributive Law A+BC = (A+B)(A+C) A(B+C) = AB+AC
Absorption Law A(A+B) = A A+AB = A
DeMorgan's Law (AB) = A+B (A+B) = A B

The Attempt at a Solution

I'm going to use lower case letters now.f=ab'cd+(abc')'+abcd'
f'=(ab'cd+(abc')+abcd')'
=(ab'cd)'(abc')''(abcd')'
=(a'+b+c'+d')(abc')(a'+b'+c'+d)
=(aa'bc'+abbc'+abc'c'+abc'd')(a'+b'+c'+d)
=(abc'+abc'+abcd')(a'+b'+c'+d)
=(abc'+abcd')(a'+b'+c'+d')
=(0+0+0+0+abc'c'+abc'c'd'+abc'd+0)
=abc'+abc'd'+abc'd
=abc'+abc'(d'+d)
=abc'+abc'(1)
=abc'
f'=abc'
f=(abc')'

--> f=a'+b'+c

Do this look correct? If so is there a shorter way to minimize it? Is there a way to minimize without using DeMorgan's theorem at the top? Thanks

Impressive. I got the same answer via the K-map.

 

[maline]

sorry my mistake, i misread the question. your answer is correct. i would expand the middle term, and then show that it dominates.

 

[Duderonimous]

Great. Thank you. I found a much quicker way.

f=ab'cd+(abc')'+abcd'
=ab'cd+a'+b'+c+abcd'
=(acd+1)b'+a'+c(1+abd')
=(1)b'+a'+c(1)
=a'+b'+c