Exercise 2: Tracing the MUX for sel=1, in0=1, in1=0 — Possible Solution
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GIVEN
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MUX formula: output = sel'.in0 + sel.in1
sel=1, in0=1, in1=0

STEP-BY-STEP TRACE
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sel' = NOT(1) = 0

First term: sel'.in0 = 0.1 = 0
Second term: sel.in1 = 1.0 = 0

output = 0 + 0 = 0

RESULT
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Output = 0, which correctly equals in1 (0) - exactly what should
happen when sel=1 selects the second input.

WHY in0 DOESN'T MATTER ONCE sel=1
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When sel=1, the first term (sel'.in0) always contains sel'=0 as a
factor - and ANY value AND-ed with 0 is always 0, regardless of
what in0 actually is (this chapter's own domination law: x.0=0).
So the first term is forced to 0 no matter what in0 is, leaving
the output entirely determined by the second term alone (sel.in1
= 1.in1 = in1) - which is precisely the intended MUX behavior:
sel=1 should always select in1, completely independent of in0.

WHY THIS WORKS AS AN ANSWER
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Both terms of the MUX formula are computed explicitly rather than
jumping straight to "sel=1 means in1," and the explanation for why
in0 becomes irrelevant is grounded in a specific named Boolean law
(domination: x.0=0) rather than just restated as "that's how a MUX
works."