Challenge 1: myLength/2 Over a Five-Element List — Possible Solution
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list_utils.pl:

myLength([], 0).
myLength([_|T], N) :- myLength(T, N1), N is N1 + 1.

Query:

    ?- myLength([a, b, c, d, e], N).
    N = 5.

Explanation:
myLength recurses down the list one element at a time -- each
recursive call strips off one head (discarded via the anonymous
variable _, since the actual value of each element is irrelevant to
computing a length) and asks for the length of the remaining tail.
The base case, myLength([], 0), finally fires once the list is fully
consumed, reporting 0. Each recursive call then adds 1 to the result
coming back from its own inner call, so the final N accumulates to 5
by the time the outermost call returns -- five elements, five
increments plus the base case's own 0.

WHY THIS WORKS AS AN ANSWER
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This uses the chapter's own myLength/2 definition exactly as written,
querying it against a genuine five-element list to confirm the
recursive relation correctly reports the real length.