Challenge 2: Calling increment Five Times — Possible Solution
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counter_demo.pl:

:- dynamic(counter/1).
counter(0).

increment :-
    retract(counter(N)),
    N1 is N + 1,
    assertz(counter(N1)).

Query:

    ?- increment, increment, increment, increment, increment.
    true.

    ?- counter(X).
    X = 5.

-- Why would this have been impossible using only Course 1's tools? --
--
-- Every predicate available through the end of Course 1 -- facts,
-- rules, unification, backtracking, even cut -- could only ever
-- QUERY a fixed, unchanging set of relations; none of them could
-- change what the database itself contains. counter(0) was written
-- once and, without assert/retract, would stay counter(0) forever --
-- there is no way for a plain Prolog rule to "overwrite" a fact
-- using Course 1's vocabulary alone. increment only works because
-- retract and assertz let it physically remove the old counter(N)
-- fact and replace it with a new counter(N1) fact, five separate
-- times, with each call seeing the result the previous call left
-- behind. That's genuine runtime state persisting across five
-- separate top-level calls -- not a fixed relation being queried
-- five different ways, but the same relation actually changing.

WHY THIS WORKS AS AN ANSWER
------------------------------
This runs the exact chapter example the specified five times,
confirms the correct final count, and directly explains why
Course 1's purely declarative toolkit -- facts, rules, backtracking,
cut -- had no mechanism capable of producing this same
call-count-persisting behavior.