Challenge 1: Concatenating [a, b] and [c, d, e] via Difference Lists — Possible Solution
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dl_demo.pl:

dl_concat(A-B, B-C, A-C).

Query:

    ?- L1 = [a, b | H1], DL1 = L1-H1,
       L2 = [c, d, e | H2], DL2 = L2-H2,
       dl_concat(DL1, DL2, Combined-H2),
       H2 = [].
    L1 = [a, b, c, d, e],
    DL1 = [a, b, c, d, e]-[],
    H1 = [c, d, e],
    L2 = [c, d, e],
    DL2 = [c, d, e]-[],
    H2 = [],
    Combined = [a, b, c, d, e].

Explanation:
DL1 represents the open list [a, b | H1], DL2 represents the open
list [c, d, e | H2]. dl_concat(DL1, DL2, Combined-H2) unifies H1 (the
first list's hole) with L2 itself, so L1 becomes [a, b, c, d, e | H2]
without walking or copying a single element -- notice L1 itself
changes as a side effect of the unification, since H1 was a shared
variable inside it all along. The final step, H2 = [], closes the
remaining open tail, turning Combined into the ordinary, fully closed
list [a, b, c, d, e].

WHY THIS WORKS AS AN ANSWER
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This builds two difference lists exactly per the chapter's own
List-Hole representation, concatenates them with dl_concat/3 as
defined, and explicitly closes the result by unifying the final hole
with [], producing the expected ordinary five-element list.