binary to NAND gate - boolean algebra

Question

I am taking a digital logic class and i am trying to multiply this binary number. I am not sure what a carry in is and a carrry out. the teachers slides are horrbile. It appears he used a truth table to do this but its confusing.

   X1X0 
 + Y1Y0
   ----
 Z2Z1Z0

I think thats how its set up! Now, for the multiplication part

    1  carry in?
110101 
X 1101
------
101011001  thats what i ended up with. Probobly, not right!

I think my truth table should look something like this: keep in mind this is not set up to my answer above

       X1X0 
     + Y1Y0
       ----
     Z2Z1Z0

       X0 Y0   Carry     Z0
       0   0     0        0
       1   0     0        1
       0   1     0        1
       1   1     1        0



  X1  Y1   Carryin            Carryout Z1 
  0   0     0                       0   0
  1   0     0                       0   1
  0   1     0                       0   1
  1   1     0                       1   0
  0   0     1                       0   1
  1   0     1                       1   0

I get confused on the x1 and y1 part It would be easier if i can see it in action and labeled what the "carry in" is and "carry out" is while its being multiplied.

would the "carry in" be the result of 1+1 and the "carry out" be the result of the next carry result?

I think after we get the truth table done with the carry in and carry out we are to use boolean algebra like:

Z1 = X1• Y1' • Carryin' + X1' • Y1• Carryin' + X1' • Y1' • Carryin + X1• Y1• Carryin 
Carryout = X1• Y1• Carryin' + X1 • Y1' • Carryin + X1' • Y1• Carryin + X1 • Y1• Carryin
Z2 = Carryout

We are to "work out the equations for the AND, OR and NOT functions using only the NAND operator." not sure how to do this!

Answer 1

NAND is simply an AND operating followed by NOT.

In terms of implementing the other boolean operations with just NAND:

  NOT a = a NAND a

a | (a NAND a) | result
--+------------+-------
0 |     1      |  OKAY
1 |     0      |  OKAY

a AND b = NOT (NOT (a AND b))
        = NOT (a NAND b)
        = (a NAND b) NAND (a NAND b)

a | b | x=(a NAND b) | (x NAND x) | result
--+---+--------------+------------+-------
0 | 0 |      1       |     0      |  OKAY
0 | 1 |      1       |     0      |  OKAY
1 | 0 |      1       |     0      |  OKAY
1 | 1 |      0       |     1      |  OKAY

a OR  b = NOT((NOT a) AND (NOT b))               # DeMorgans Law
        = NOT((a NAND a) AND (b NAND b))
        = NOT(NOT ((a NAND a) NAND (b NAND b)))
        = (a NAND a) NAND (b NAND b)

a | b | x=(a NAND a) | y = (b NAND b) | (x NAND y) | result
--+---+--------------+----------------+------------+-------
0 | 0 |       1      |        1       |      0     |  OKAY
0 | 1 |       1      |        0       |      1     |  OKAY
1 | 0 |       0      |        1       |      1     |  OKAY
1 | 1 |       0      |        0       |      1     |  OKAY