# Convert Truth Table to Boolean AND-OR Expression

• Mar 22nd 2014, 02:16 PM
lamentofking
Convert Truth Table to Boolean AND-OR Expression
Hello MHF. I am stuck on a problem that goes like this:

Write the boolean AND-OR expression for the following:

function x in Figure 10.17

Attachment 30489

I don't understand how to get a boolean expression from that truth table. Are true and false reversed for this problem meaning, True = 0 and False = 1? And do I just write down the expressions everywhere x = 1? So for example where a, b, and c are 0 and x=1, the expression for that part would be abc? Here's what I have so far:

abc + abc' + ab'c + a'bc + a'bc' + (abc)'

Is that right?
• Mar 22nd 2014, 11:12 PM
romsek
Re: Convert Truth Table to Boolean AND-OR Expression
you certainly can just list the combinations that result in 1 in your truth table but I suspect they want you to simplify that.

The first four entries can be seen to be

\$a'(bc)'\$

The second four can be expressed as

\$ab'+ac\$

so the total expression is

\$a'(bc)'+ab'+ac\$ and this minimizes to

\$(ac + a'c'+b')\$
• Mar 23rd 2014, 11:52 AM
lamentofking
Re: Convert Truth Table to Boolean AND-OR Expression
Quote:

Originally Posted by romsek
you certainly can just list the combinations that result in 1 in your truth table but I suspect they want you to simplify that.

The first four entries can be seen to be

\$a'(bc)'\$

The second four can be expressed as

\$ab'+ac\$

so the total expression is

\$a'(bc)'+ab'+ac\$ and this minimizes to

\$(ac + a'c'+b')\$

Can you explain how the first four entries can be seen as a'(bc)' ? What I got for the first four entries in the truth table (starting at the top) are abc + abc' + ab'c + a(bc)'. I simplified it as such:

abc + abc' + ab'c + a(bc)' = ab(c + c') + ab'(c + c')
= (c + c') + (ab + ab')
= 1 + a(b + b')
= 1 + a(1)
= 1 + a
= a

I didn't try the second four since I didn't understand the first four.
• Mar 23rd 2014, 03:32 PM
romsek
Re: Convert Truth Table to Boolean AND-OR Expression
Quote:

Originally Posted by lamentofking
Can you explain how the first four entries can be seen as a'(bc)' ? What I got for the first four entries in the truth table (starting at the top) are abc + abc' + ab'c + a(bc)'. I simplified it as such:

abc + abc' + ab'c + a(bc)' = ab(c + c') + ab'(c + c')
= (c + c') + (ab + ab')
= 1 + a(b + b')
= 1 + a(1)
= 1 + a
= a

I didn't try the second four since I didn't understand the first four.

the first four entries all correspond to a' as a is 0 for all four.

the relationship of b and c in those first four entries is (bc)'. That should be pretty clear.

Combine those two facts to get that the first four entries correspond to a'(bc)'

000 - 1(00)' - 1(0)' - 11 - 1
001 - 1(01)' - 1(0)' - 11 - 1
010 - 1(10)' - 1(0)' - 11 - 1
011 - 1(11)' - 1(1)' - 10 - 0
• Mar 23rd 2014, 05:29 PM
lamentofking
Re: Convert Truth Table to Boolean AND-OR Expression
Quote:

Originally Posted by romsek
the first four entries all correspond to a' as a is 0 for all four.

the relationship of b and c in those first four entries is (bc)'. That should be pretty clear.

Combine those two facts to get that the first four entries correspond to a'(bc)'

000 - 1(00)' - 1(0)' - 11 - 1
001 - 1(01)' - 1(0)' - 11 - 1
010 - 1(10)' - 1(0)' - 11 - 1
011 - 1(11)' - 1(1)' - 10 - 0

Sorry, my book doesn't do a good job of explaining this. Is that a rule to convert to AND-OR expression that with a value like 000 you pull out a 1 and take the complement of the last two numbers and simplify? I can do that pattern:

000 - 1(00)' - 1(0)' - 11 - 1
001 - 1(01)' - 1(0)' - 11 - 1
010 - 1(10)' - 1(0)' - 11 - 1
011 - 1(11)' - 1(1)' - 10 - 0

for the second 4 as well?