1. ## valid arguement?

p goes to q
not p
therefore not q

I think this is invalid because while not q could be the case q could also be true since p is false. Is this sound logic?

p-->(q-->r)

q--> (p-->r)

therefore (p or q) --> r

I know this is invalid because I have the answer but I do not understand why??? Could someone please explan this? Thank you

2. Hello, frostking2!

Are you supposed to talk your way through these?
What methods have you been taught?
These can be solved with truth tables . . .

$\begin{array}{c}p \to q\\
\sim p \\ \hline
\therefore\;\sim q\end{array}$
This is reasoning by the Inverse . . . which is invalid.

Or we can run this through a truth table: . $\left[(p \to q)\:\wedge \sim p\right] \to \:\sim q$

$\begin{array}{cc} & p \to (q\to r) \\
& q \to (p \to r) \\ \hline
\therefore & (p \vee q) \to r \end{array}$
Make a truth table for: . $\bigg(\left[p\to (q\to r)\right] \wedge \left[q\to(p\to r)\right]\bigg) \:\to\:\left[(p \vee r) \to r\right]$

3. for frostking2's 2nd question, is there an easier way to do this? cause i stink at making truth tables, and constantly make mistakes on them...
Have a great one, and thanks in advance...