Challenge 2: Summing a Longer Expression List — Possible Solution
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expr_demo.pl:

expr(Sum) --> term(T), expr_rest(T, Sum).

expr_rest(Acc, Sum) -->
    [plus], term(T),
    { Acc1 is Acc + T },
    expr_rest(Acc1, Sum).
expr_rest(Sum, Sum) --> [].

term(T) --> [T], { number(T) }.

Query:

    ?- phrase(expr(Sum), [5, plus, 10, plus, 15, plus, 20]).
    Sum = 50.

Explanation:
expr/1 first matches a single term (5), then hands off to
expr_rest(5, Sum) to consume the rest of the list. expr_rest
recursively matches [plus], term(T) three more times -- once for
each remaining plus/number pair -- computing a running total via
{ Acc1 is Acc + T } at each step: 5+10=15, then 15+15=30, then
30+20=50. Once the input list is fully consumed, expr_rest's second
clause (expr_rest(Sum, Sum) --> [].) matches the empty remaining
input and unifies the final accumulator directly with Sum, giving the
correct total of 50 regardless of how many plus-separated terms the
list actually contains.

WHY THIS WORKS AS AN ANSWER
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This runs the chapter's own grammar unchanged against a longer,
four-term input list, confirming the recursive accumulator pattern in
expr_rest/2 correctly scales beyond the chapter's original three-term
example without any modification to the grammar itself.