Discrete Math help?

Question

Here PLEASE http://www.physicsforums.com/showthread.php?t=153872 ?

Answer 1

(p → (q → r)) ↔ ((p ∧ q) → r)

Using p ↔ q = (p → q) ∧ (q → p):

((p → (q → r)) → ((p ∧ q) → r)) ∧ (((p ∧ q) → r) → (p → (q → r)))

Using p → q = ¬p ∨ q at each innermost expression:

((p → (¬q ∨ r)) → (¬(p ∧ q) ∨ r)) ∧ ((¬(p ∧ q) ∨ r) → (p → (¬q ∨ r)))

And again:

((¬p ∨ (¬q ∨ r)) → (¬(p ∧ q) ∨ r)) ∧ ((¬(p ∧ q) ∨ r) → (¬p ∨ (¬q ∨ r)))

And again:

((¬(¬p ∨ (¬q ∨ r)) ∨ (¬(p ∧ q) ∨ r))) ∧ ((¬(¬(p ∧ q) ∨ r) ∨ (¬p ∨ (¬q ∨ r))))

That may be as far as you need to go, but I hate parentheses, so I'd rather distribute and use Demorgan's as necessary.

((¬(¬p ∨ ¬q ∨ r) ∨ (¬p ∨ ¬q ∨ r))) ∧ ((¬(¬p ∨ ¬q ∨ r) ∨ (¬p ∨ ¬q ∨ r)))

((p ∧ q ∧ ¬r) ∨ (¬p ∨ ¬q ∨ r)) ∧ ((p ∧ q ∧ ¬r) ∨ (¬p ∨ ¬q ∨ r))

Both sides appear to have ended up the same, so I think you can just reduce to:

(p ∧ q ∧ ¬r) ∨ (¬p ∨ ¬q ∨ r)

If I didn't screw up.

Answer 2

I get that the two statements on each side are the same.
p->(q->r)
Using the fact that p->q is equivalent to ~pVq, we have
~pV(q->r)
Using the same fact again,
~pV(~qVr)
Since they are all or, you can just drop the parentheses.
~p V ~q V r

Now for the other one.
(p&q)->r
~(p&q) V r
~p V ~q V r

If you let t = ~p V ~q V r, then you have
t <--> t

Double check what I did.