Question
Hi, I just need help getting started on this..
Prove the last four rows of identities (actually there are eight identities in the four last lines of the theorems - one for AND, and one for OR) To do these do a truth table for both sides of the equal sign. If you are comparing A and B you have two inputs so you need four rows in each truth table. If you have A, B, and C you have three inputs and you will need eight rows in each truth table. If the truth tables on both sides of the equal sign of the identity are exactly the same you have proved the theorem. If they differ in any way then the identity is proved wrong (but it probably means you made a mistake since others have proved these identities right). Always use counting in digital to set up the options for the inputs. If two inputs the inputs should be arranged like this: For zero, one, two, and three. If there are three inputs the inputs should be arranged like this: For zero, one. two. three, four, five, six. and seven. Then you can be sure you have covered all possibilities. There are eight identities you need to prove by doing this, so there will be sixteen truth tables. Be sure to label each truth table so I know which truth table goes with which part of the problem. These are the identities for Associative, Distributive, Absorption, and De Morgan's Law
Explanation / Answer
1) associative (a+b)+c = a+(b+c)
a b c (a+b)+c
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1
a b c a+(b+c)
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1
a b c a. (b+c)
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
a b c a.b+a.c
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
therefore a. (b+c) = a.b+a.c
a b c a+(b.c)
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
a b c (a+b) .(a+c)
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
hence verified
