Exercise 2: The Dual of the Identity Law x.1=x — Possible Solution
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STEP 1: APPLY THE DUALITY PRINCIPLE
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Per this chapter's own duality principle, the dual of a law is
found by swapping every + with . , and every 0 with 1.

Starting law: x . 1 = x

Swap . for +: x + 1 = x  (not finished yet - still need to swap
the constant)
Swap 1 for 0: x + 0 = x

Dual: x + 0 = x

STEP 2: VERIFY IT'S ACTUALLY TRUE, FOR BOTH VALUES OF x
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x=0: x+0 = 0+0 = 0. Does this equal x (which is 0)? Yes.
x=1: x+0 = 1+0 = 1. Does this equal x (which is 1)? Yes.

RESULT
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The dual of x.1=x is x+0=x, and it checks out as true for both
possible values of x - exactly this chapter's own stated identity
law's own second half, confirming the duality principle correctly
predicted a true law without needing to guess or check a truth
table first.

WHY THIS WORKS AS AN ANSWER
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The dual is derived by applying the exact swap rule this chapter
states (both the operator swap AND the constant swap, not just
one of the two), and the claim is then independently verified with
a truth table rather than just trusted because duality "should"
work.