Exercise 2: POS Form for XOR — Possible Solution
====================================================================

GIVEN
------------------------------
Same XOR function: f(x,y)=1 for (0,1) and (1,0); f(x,y)=0 for
(0,0) and (1,1)

STEP 1: IDENTIFY THE 0-ROWS AND THEIR MAXTERMS
------------------------------
Row x=0,y=0: maxterm is x+y (only 0 when x=0 AND y=0)
Row x=1,y=1: maxterm is x'+y' (only 0 when x=1 AND y=1)

STEP 2: AND THE MAXTERMS TOGETHER
------------------------------
POS form: (x+y)(x'+y')

STEP 3: VERIFY AGAINST THE ACTUAL XOR TABLE
------------------------------
x=0,y=0: (x+y)=0+0=0, (x'+y')=1+1=1. POS = 0.1 = 0. XOR(0,0)=0.
  Match.
x=0,y=1: (x+y)=0+1=1, (x'+y')=1+0=1. POS = 1.1 = 1. XOR(0,1)=1.
  Match.
x=1,y=0: (x+y)=1+0=1, (x'+y')=0+1=1. POS = 1.1 = 1. XOR(1,0)=1.
  Match.
x=1,y=1: (x+y)=1+1=1, (x'+y')=0+0=0. POS = 1.0 = 0. XOR(1,1)=0.
  Match.

RESULT
------------------------------
POS form of XOR: (x+y)(x'+y') - verified to match the real XOR
truth table for all 4 input combinations, exactly matching
Exercise 1's own SOP form (x'y + xy') on every input, just built
via the opposite (maxterm/AND) construction.

WHY THIS WORKS AS AN ANSWER
------------------------------
The maxterm for each 0-row is derived individually (using plain,
not complemented, variables for the ones that were 0 in that row -
the maxterm rule is the mirror image of the minterm rule), AND-ed
together per this chapter's own POS procedure, and every one of the
4 possible inputs is checked against the actual XOR definition.