Why is p ∧ q ⇒ r true when p is true and q is false? I'm taking Intro to Logic on Coursera.  One of the exercises has this:

Consider a truth assignment in which p is true, q is false, r is true.
  Use this truth assignment to evaluate the following sentences.

The answer key says $p ∧ q ⇒ r$ is true, but I don't understand why.  If I read it correctly, it says "If TRUE and FALSE then TRUE."  
I thought TRUE and FALSE should imply FALSE because the two propositions are different.  Can anyone explain?

If we have predicates $p$ and $q$ then $p\implies q$ is true when either $p$ is false or $q$ is true (or both). That is,
$$(p\implies q)\iff (\lnot p\lor q)$$
The implication will not hold only when $p$ is true and $q$ is false. That is, $p\implies q$ is false when $p\land \lnot q$ is true.
So in your example, $p$ is true and $q$ is false, so $p\land q$ is false. Thus, $p\land q\implies r$ is indeed true (since the "if" part of the implication is false). We see this as
$$(p\land q\implies r)\iff (\lnot(p\land q)\lor r)$$
