Challenge 2: append/3 Forward and Backward — Possible Solution
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Forward (concatenation):

    ?- append([1, 2], [3, 4], X).
    X = [1, 2, 3, 4].

Backward (finding every split):

    ?- append(X, Y, [1, 2, 3, 4]).
    X = [], Y = [1, 2, 3, 4] ;
    X = [1], Y = [2, 3, 4] ;
    X = [1, 2], Y = [3, 4] ;
    X = [1, 2, 3], Y = [4] ;
    X = [1, 2, 3, 4], Y = [].

Explanation:
The forward query supplies both pieces (X and Y are already concrete
lists) and asks append/3 to compute their concatenation directly --
ordinary, expected behavior. The backward query instead supplies only
the FINAL, already-concatenated list, leaving X and Y as unbound
variables -- append/3's own single definition, run through
backtracking, searches for every possible pair of lists that could
have concatenated to produce [1, 2, 3, 4], finding all five genuine
splits (including the two "empty piece" edge cases) without any
separate "reverse append" predicate ever being written.

WHY THIS WORKS AS AN ANSWER
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This runs the exact same append/3 predicate in both the chapter's own
forward and backward directions, showing all five real splits the
backward direction finds via backtracking, confirming append/3's
genuine bidirectionality.