# Sorry everyone!!

• Aug 9th 2006, 07:37 AM
kodirl
Sorry everyone!!
I'm sorry that i'm asking so many q's it's just that im repeating my discrete maths exam nxt wk.Im studying Cs and need to pass it to get to nx yr..So here is another Q.The nand operator is important in logic circuit theory, and is defined by

p nand q=¬(p^q)
By means of truth tables or otherwise show that;
(i)¬p=(p nand p)
(ii)pVq|=(p nand p) nand (q nand p)

preferably truth tables...
Thanks,
Kod
• Aug 9th 2006, 12:07 PM
ThePerfectHacker
Quote:

Originally Posted by kodirl
(i)¬p=(p nand p)

I believe that this is wrong.
p nand q= -(p^q)=-p v -q (By de Morgan's laws).
You cannot conclude that,
-p v -q = -p
Unless -q is false, i.e. q is true.

Quote:

Originally Posted by kodirl
(ii)pVq=(p nand p) nand (q nand p)

(p nand p) nand (q nand q) = -[(p nand p) ^ (q nand q)]
=-(p nand p) v -(q nand q)=-[-(p ^ p)] v -[-(q ^ q)]
=(p^p) v (q^q)=p v q
• Aug 9th 2006, 07:43 PM
Soroban
Hello, Kid!

Here's the first one . . .

Quote:

By means of truth tables or otherwise show that:
. . . $(1)\;\sim p \;\longleftrightarrow \;(p\text{ nand }p)$

The truth values for nand are:
Code:

    p | q | p nand q     - + - + - - - - -     T | T |    F     T | F |    T     F | T |    T     F | F |    T

For (1) we have:
Code:

    p | ~p  <--->  (p nand p)     - + - - - - - - - - - - -     T |  F    T    T  F  T     F |  T    T    F  T  F               ↑