# CMPSCI 250: Introduction to Computation

### David Mix Barrington

### Spring, 2005

# Homework #1 Questions and Answers

#### Question 1.3, 8 February 2005

I'm not sure what I'm being asked to do in Problem 1.8.3. Should I prove
"(P∧¬Q)→0" by cases?

No, you are asked to give a direct proof of "P→Q", which
means that you should assume P and prove Q. To make *this* proof a proof by
cases, you need to pick some proposition R and then both (1) assume "P∧R" and prove
Q, and (2) assume "P∧not;R" and prove Q. You are given that you know how to prove
0 from "P∧¬Q" -- using this proof and other valid rules, tell me how you can
prove both (1) and (2) with the R of your choice.

#### Question 1.2, 3 February 2005

I am confused by the solution to Exercise 1.8.1. In the first line,
how can we assume "0∧p" when it is obviously false? And don't the
second and fourth lines contradict each other?

It is true that no assignment of truth value to p can make "0∧p" true.
But we can still assume it for the purposes of making a direct proof of the
implication "(0∧p)→0", which is what we want to prove here. The
second line of the proof is part of a "fantasy world" where "0∧p" is
assumed to be true. By showing that "0" is true in this fantasy world, we
complete the direct proof. Line 4, on the other hand, is in the real world
where "0" is false. So there is no conflict with line 2, which has an
assumption in force.

#### Question 1.1, 31 January 2005

In Problem 1.5.2, the "and" symbol looks a little strange.
Can I take the "and" of two sets?

No, you can't. The "and" (∧) is a mistake and should be an
"intersection" (∩) symbol.

Can I show a counterexample with Venn diagrams in Problem 1.5.2?

As with diagrams in geometry, Venn diagrams can illustrate a proof or
an example but don't really give the example or proof itself. A good
way to use Venn diagrams in this problem is as follows. Suppose the
statement were "A ∩ B = A ∩ C". You draw a Venn diagram and find
that the region "A ∩ B ∩ C-bar" is in one set but not the other.
So any three sets, where some element x is in A and B but not in C, will work.
The simplest example is A = B = {x} and C = ∅ so A ∩ B = {x}
and A ∩ C = ∅ and the statement isn't true.

Last modified 8 February 2005