Exercise 1: Truth Table for (p v q) ^ ~p — Possible Solution
====================================================================

BUILDING THE TRUTH TABLE
------------------------------
Working through all four combinations of p and q, computing (p v q)
first, then ~p, then combining the two with AND:

  p | q | p v q | ~p | (p v q) ^ ~p
  --|---|-------|----|--------------
  T | T |   T   | F  |      F
  T | F |   T   | F  |      F
  F | T |   T   | T  |      T
  F | F |   F   | T  |      F

WORKING THROUGH EACH ROW
------------------------------
Row 1 (p=T, q=T): p v q = T (at least one is true). ~p = F (p is
true). T AND F = F.

Row 2 (p=T, q=F): p v q = T (p is true). ~p = F. T AND F = F.

Row 3 (p=F, q=T): p v q = T (q is true). ~p = T (p is false). T AND T
= T.

Row 4 (p=F, q=F): p v q = F (neither is true). ~p = T. F AND T = F.

THE RESULT: ONLY TRUE IN ROW 3
------------------------------
The compound proposition is true in exactly one case: p is false and
q is true. In every other combination, it comes out false.

WHAT THIS ACTUALLY CAPTURES IN PLAIN ENGLISH
------------------------------
The expression is true exactly when "q is true and p is false" -
in other words, exactly the same condition as ~p ^ q (q but not p).
This can be checked directly: ~p ^ q gives ~p = F,F,T,T and q =
T,F,T,F across the same four rows, so ~p ^ q = F,F,T,F - identical to
the (p v q) ^ ~p column computed above. Both expressions are logically
equivalent, per this chapter's own definition of equivalence (matching
truth tables row by row).

WHY THIS WORKS AS AN ANSWER
------------------------------
It builds the truth table one column at a time rather than jumping
straight to the answer, explains the reasoning behind each row's own
result, and translates the single true row into a plain-English
description ("q but not p"), verified by comparing against the
simpler equivalent expression's own truth table.