Exercise 3: Is yz Really Redundant in xy + xz + yz? — Possible Solution
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THE COLLEAGUE'S CLAIM
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That xy + xz + yz can be safely reduced to just xy + xz, with yz
dropped as redundant.

STEP 1: CHECK ALL 8 INPUT COMBINATIONS
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x=0,y=0,z=0: full(xy+xz+yz)=0, reduced(xy+xz)=0. Match.
x=0,y=0,z=1: full=0, reduced=0. Match.
x=0,y=1,z=0: full=0, reduced=0. Match.
x=0,y=1,z=1: full=1, reduced=0. MISMATCH.
x=1,y=0,z=0: full=0, reduced=0. Match.
x=1,y=0,z=1: full=1, reduced=1. Match.
x=1,y=1,z=0: full=1, reduced=1. Match.
x=1,y=1,z=1: full=1, reduced=1. Match.

STEP 2: EXAMINE THE MISMATCH
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At x=0, y=1, z=1: the full expression (xy+xz+yz) evaluates to 1,
because yz = 1.1 = 1 (both x-containing terms, xy and xz, are 0
here since x=0). The reduced expression (xy+xz) evaluates to 0,
since both its terms need x=1 to ever be 1, and x=0 here.

RESULT
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The colleague is WRONG. Dropping yz is not safe - it changes the
function's output for the specific case x=0,y=1,z=1, where yz is
the ONLY term keeping the overall expression true. Every one of
the three terms in xy+xz+yz is genuinely necessary; none of them
can be removed without changing the function.

WHY THIS SEEMS PLAUSIBLE BUT ISN'T
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It's tempting to assume any term built entirely from "already
present" variables (y and z both appear in the other two terms
too) must be redundant - but that reasoning only works if every
case yz covers is ALSO covered by xy or xz. Here it isn't: yz is
the only term that can be true when x=0, since the other two terms
both require x=1.

WHY THIS WORKS AS AN ANSWER
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Rather than trusting or dismissing the colleague's claim on
intuition, all 8 input combinations are checked explicitly, the
single mismatching case is examined to explain WHY the reduced
form fails there specifically, and the underlying reasoning error
(assuming a term with "already-used" variables must be redundant)
is named directly rather than left implicit.