Challenge 2: Progressively Refining a Partially-Bound Term — Possible Solution
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SWI-Prolog session:

    ?- pair(X, Y) = pair(f(Z), 10), Z = 42.
    X = f(42),
    Y = 10,
    Z = 42.

Explanation:
The first goal, pair(X, Y) = pair(f(Z), 10), unifies X with f(Z) --
a compound term that ITSELF still contains the unbound variable Z --
and unifies Y directly with 10. At this point X is only PARTIALLY
known: it's "f of something," but that something isn't concrete yet.
The second goal, Z = 42, then unifies Z with 42. Because X was bound
to f(Z) (not a copy of Z's value at that moment, but a genuine
reference to the same variable), binding Z retroactively completes
X's own value too -- X now reports as f(42), fully resolved, even
though it was only ever unified once, before Z had a value at all.

WHY THIS WORKS AS AN ANSWER
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This reproduces the chapter's own f(X, Y) = f(g(Z), 5), Z = 3 example
with a renamed structure, confirming the exact "partially bound, later
completed" behavior the chapter names as having no Haskell pattern-
matching equivalent.