Determine if XNOR is distributive over addition XNORa b is 1

Determine, if XNOR is distributive over addition. [XNOR(a, b) is 1 when both a and b are 1 or both a and b are 0.]

Solution

Think about it...

sa = 1 ca = 1
sb = 1 cb = 1
a = sa + ca = 2
b = sb + cb = 2

(a | b) = 2
(a & b) = 2
(sa | sb) + (ca | cb) = (1 | 1) + (1 | 1) = 1 + 1 = 2 # Correct or Coincidence?
(sa & sb) + (ca & cb) = (1 & 1) + (1 & 1) = 1 + 1 = 2 # Correct or Coincidence?
Let\'s try some other values:

sa = 1001 ca = 1 # Binary
sb = 0100 cb = 1
a = sa + ca = 1010
b = sb + cb = 0101
(a | b) = 1111
(a & b) = 0000
(sa | sb) + (ca | cb) = (1001 | 0101) + (1 | 1) = 1101 + 1 = 1110 # Oh dear!
(sa & sb) + (ca & cb) = (1001 & 0101) + (1 & 1) = 0001 + 1 = 2 # Oh dear!

So, its proof by 4-bit counter example that you cannot distribute AND or OR over addition.

What about \'>>>\' (unsigned or logical right shift). Using the last example values, and r = 1:

sa = 1001
ca = 0001
sa >>> 1 = 0101
ca >>> 1 = 0000
(sa >>> 1) + (ca >>> 1) = 0101 + 0000 = 0101
(sa + ca) >>> 1 = (1001 + 0001) >>> 1 = 1010 >>> 1 = 0101 # Coincidence?
Let\'s see whether that is coincidence too:

sa = 1011
ca = 0001
sa >>> 1 = 0101
ca >>> 1 = 0000
(sa >>> 1) + (ca >>> 1) = 0101 + 0000 = 0101
(sa + ca) >>> 1 = (1011 + 0001) >>> 1 = 1100 >>> 1 = 0110 # Oh dear!
Proof by counter-example again.

So logical right shift is not distributive over addition either.