What is the truth set of this logic? I am trying to solve this question.  From the prompt, I figure:
$S = \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19\}$
$p = \{1,  3,  5,  7,  9\}$
$q = \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$
I appears $p → q$ is $\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$, which is $q$.  How come my answer is wrong?  Am I missing something?
 A: You're missing the numbers above $15$ where $p$ and $q$ are both false.
If the premise is false, the implication is considered to be true.
Note that, 'being odd and smaller than $10$' implies 'being smaller or equal than $15$', so this implication should be fully true. 
A: It would be clear if we write it as following:
$$P(n)\equiv n\in\{1,3,5,7,9\},~Q(n)\equiv n\in[1,15]$$
$$P(n)\to Q(n)\equiv\neg P(n)\lor Q(n)\equiv n\in S\setminus\{1,3,5,7,9\}\cup[1,15]$$
$$S\setminus\{1,3,5,7,9\}\cup[1,15]=S$$
It's because $p\to q$ is still true when $p$ is false, and the set which make $p$ false is: $$S\setminus\{1,3,5,7,9\}=\{2,4,6,8\}\cup[10,19]$$
Combine this with $[1,15]$ which will give us the whole set $S$.
A: We have two cases:
$$p=False \Longrightarrow (p \rightarrow q) = True \Longrightarrow p^c \subseteq Truth(p \rightarrow q) $$
$$p=True \Longrightarrow (p \rightarrow q) \sim q \Longrightarrow (p \cap q) \subseteq Truth(p \rightarrow q)$$
So we have:
$$Truth(p \rightarrow q) = p^c \cup (p \cap q) = (q-p)^c$$
In your case:
$$p \subseteq q \Longrightarrow Truth(p \rightarrow q) = S \Longrightarrow (p \rightarrow q) \sim T$$
