A couple of Natural Deduction proofs I have two proofs that I can't figure out how to get started on.
a) q ├ (p ∨ ¬p) & q
b) p ∨ q, p→¬q Ⱶ (p→q)→(q ∧ ¬p)

for q ├ (p ∨ ¬p) & q  I only assumed that I might try to prove it indirectly:
1)q     (Assumption)
2)show (p ∨ ¬p) & q
3)| show p ∨ ¬p
4)||¬(p ∨ ¬p)     (Assume for Indirect Proof)
5)||???
And really thats as far as I can figure out how to get. I don't know what I could do with line 4 or with q to get much else...My other idea was or extraction on p ∨ ¬p but I don't see a way to show p→¬p to make that work...

For p ∨ q, p→¬q Ⱶ (p→q)→(q ∧ ¬p) I thought a good first step would be or extraction on p ∨ q but cant get past the first step of that one...

1)p ∨ q     (Assumption)
2)p→¬q     (Assumption)
3)(p→q)→(q ∧ ¬p)
4)|show p → q
5)||q      (Assume for Conditional Proof)
6)||???

So yeah. I guess I'm a bit out of my league on both of these. Any assistance at all would be helpful for I've been staring at these two problems for close to an hour.
 A: (a) It is a standard bit of "bookwork" to prove, from no assumptions, $(p \lor \neg p)$ and you should make sure you know how to do it in your favourite natural deduction system. Here's the strategy.
What else can you do but assume the opposite and aim for a contradiction, so you start by making the temporary assumption

$\quad\quad | \quad \neg(p \lor \neg p)$

Now what can you do? No standard rules can be used to extract anything from this, so we'll need to make another assumption to proceed. but what? Well, sensibly, we don't want to introduce a new variable. So something involving just $p$. Let's try the simplest option (because it so quickly leads to a contradiction!), so we get

$\quad\quad | \quad \neg(p \lor \neg p)$
$\quad\quad | \quad\quad | \quad p$
$\quad\quad | \quad\quad | \quad (p \lor \neg p)$
$\quad\quad | \quad\quad | \quad \bot$

where we used or-introduction and immediately hit contradiction! So we can now use reductio and discharge the supposition. So to continue in the same vein ...

$\quad\quad | \quad \neg p$
$\quad\quad | \quad (p \lor \neg p)$
$\quad\quad | \quad \bot$

And lo! Our original supposition has now itself been reduced to absurdity and we can discharge it to get

$\neg\neg(p \lor \neg p)$

which classically entails

$(p \lor \neg p)$.

Exercises. Restyle this proof into your favourite natural deduction system. Then plug in the result to use it to complete the given exercise.
(b) The second proof should write itself if you think strategically. You know that you have to start like this:

$ (p \lor q)$
$ (p \to \neg q)$
$\quad \quad | \quad (p \to q)$

That's the two given premiss, and since we want to prove $(p \to q) \to (q \land \neg p)$ we use the standard tactic for proving a conditional result, i.e. assume the antecedent, and aim to derive the consequent, ok? So we make the temporary additional assumption $p \to q$, and aim to prove $q \land \neg p$.
Ok now note that have both $p \to q$ and $p \to \neg q$. Ah hah! That means $p$ leads to a contradiction and must be false! So .... let's use that!

$ (p \lor q)$
$ (p \to \neg q)$
$\quad \quad | \quad (p \to q)$
$\quad \quad | \quad\quad | \quad p $
$\quad \quad | \quad\quad | \quad q $
$\quad \quad | \quad\quad | \quad \neg q$
$\quad \quad | \quad\quad  | \quad \bot$
$\quad \quad | \quad\neg p $

So that's half what we want! But now we have $\neg p$ and $p \lor q$ and that gives us $q$ by disjunctive syllogism, which will either be a basic rule of your natural deduction system, or a standard bit of bookwork to get by your rules. So you just need to continue, filling in as necessary ...

$\quad \quad | \quad\vdots $
$\quad \quad | \quad q $
$\quad \quad | \quad (q \land \neg p)$
$(p \to q) \to (q \land \neg p)$

where at the last stage we discharge that assumption back at line 3 by conditional proof. All done!
