Exercise 1: SOP Form for XOR — Possible Solution
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GIVEN
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f(x,y) = 1 only when x=0,y=1 or x=1,y=0 (this is XOR)

STEP 1: IDENTIFY THE 1-ROWS AND THEIR MINTERMS
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Row x=0,y=1: minterm is x'y (only 1 when x=0 AND y=1)
Row x=1,y=0: minterm is xy' (only 1 when x=1 AND y=0)

STEP 2: OR THE MINTERMS TOGETHER
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SOP form: x'y + xy'

STEP 3: VERIFY AGAINST THE ACTUAL XOR TABLE
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x=0,y=0: x'y = 1.0=0, xy' = 0.1=0. SOP = 0+0 = 0. XOR(0,0)=0. Match.
x=0,y=1: x'y = 1.1=1, xy' = 0.0=0. SOP = 1+0 = 1. XOR(0,1)=1. Match.
x=1,y=0: x'y = 0.0=0, xy' = 1.1=1. SOP = 0+1 = 1. XOR(1,0)=1. Match.
x=1,y=1: x'y = 0.1=0, xy' = 1.0=0. SOP = 0+0 = 0. XOR(1,1)=0. Match.

RESULT
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SOP form of XOR: x'y + xy' - verified to match the real XOR truth
table for all 4 input combinations.

WHY THIS WORKS AS AN ANSWER
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The minterm for each 1-row is derived individually (correctly
complementing the variable that's 0 in that row), OR-ed together
per this chapter's own SOP procedure, and every one of the 4
possible inputs is checked against the actual XOR definition rather
than just asserting the formula is correct.