Deductive agruments in finite math: wrong answer in book?

[Back2College]

Homework Statement

Identify the following as a valid or an invalid argument.

p → q
q ∧ r
--------------
∴ ~r → ~p

Homework Equations

N/A

The Attempt at a Solution

Truth table values:

(a) p → q TTFFTTTT
(b) q ∧ r TFFFTFFF
(c) a ∧ b TFFFTFFF
(d) ~r → ~p TFTFTTTT
(e) c → d TTTTTTTT

Since (e) is true in every case, the argument is valid. However, the answer in the back of the book says invalid.

Am I doing something wrong, or is the answer in the book wrong?

 

[haruspex]

Am I doing something wrong, or is the answer in the book wrong?

I get your answer, but just from reasoning. You are given r and q are true, so the predicate not r leads to contradiction. A false predicate implies anything at all.

 

[Mark44]

a) p → q TTFFTTTT

It's hard to tell what you mean by this and the other logical expressions you show.
For a truth table for $p \Rightarrow q$, you need three columns: one for p, one for q, and one for the implication, like this:
$\begin{array} {ccc} \\ p & q & p \Rightarrow q \\ T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \end{array}$
The only pair of values for p and q for which the implication is false is when p is true and q is false. In all other cases, the implication is true.

Edit: Miswrote something in my truth table, but now fixed.

 

[haruspex]

It's hard to tell what you mean by this and the other logical expressions you show.
For a truth table for $p \Rightarrow q$, you need three columns: one for p, one for q, and one for the implication, like this:
$\begin{array} {ccc} \\ p & q & p \Rightarrow q \\ T & T & T \\ T & F & F \\ F & T & T \\ F & F & F \end{array}$
The only pair of values for p and q for which the implication is false is when p is true and q is false. In all other cases, the implication is true.

Shouldn't the last line in the table read FFT?

 

[Mark44]

Shouldn't the last line in the table read FFT?

Yes -- now fixed.