Exercise 1: Translating "Every Function in This Module Has a Docstring" — Possible Solution
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CHOOSING THE DOMAIN AND PREDICATES
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Domain: all functions (the broad universe of discourse, not just
this module's own functions).

Two predicates are needed:
  InModule(x)    — "x is a function defined in this module"
  HasDocstring(x) — "x has a docstring"

THE TRANSLATION
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  for all x, ( InModule(x) -> HasDocstring(x) )

In words: for any function x at all, IF it's in this module, THEN it
has a docstring. This matches this chapter's own "every P is a Q"
pattern exactly, with P = InModule and Q = HasDocstring.

WHY AND (^) WOULD BE THE WRONG CONNECTIVE HERE
------------------------------
Writing "for all x, (InModule(x) ^ HasDocstring(x))" instead would
claim that EVERY function in the entire universe of functions - not
just this module's own - is simultaneously in this module AND has a
docstring. That's a wildly different, and obviously false, statement:
it would mean no function exists anywhere outside this one module,
since every function would have to satisfy InModule(x).

WHY THE -> VERSION IS THE ONLY ONE THAT MATCHES THE ENGLISH MEANING
------------------------------
The original English sentence only makes a claim about functions that
happen to be in this module - it says nothing at all about functions
elsewhere. The -> version captures exactly that conditional scope:
for any function that isn't in this module, InModule(x) is false, and
per Chapter 2's own truth table, an implication with a false premise
is automatically true regardless of HasDocstring(x) - meaning
functions outside the module simply don't affect whether the overall
statement holds at all. This is precisely the intended meaning: the
claim is only actually "tested" against functions that are in the
module.

WHY THIS WORKS AS AN ANSWER
------------------------------
It defines an explicit domain and two named predicates, produces the
translation using this chapter's own "every P is Q" pattern, and
explains concretely why the alternative (^) reading would assert a
much stronger and false claim about every function everywhere, not
just the ones actually in scope.