i need hints to this logic problem?

Question

the problem is , simplify the following statement :

(p v r ) --> [ (q v (~r) ) --> ( (~p) --> r ) ]
what i did was , first the right hand side:
[ (q v (~r) ) --> ( p v r ) ] ( because ~p-->r is equivalent to p v r)
[ ~ (q v (~r) ) v ( p v r ) ] ( here i used the same logic rule)
[ (~ q ^ r ) v ( p v r ) ]
then
( p v r ) --> [ (~ q ^ r ) v ( p v r ) ]
then i used the equivalence p --> q == ~ p v q
in which p is ( p v r ) and q is [ (~ q ^ r ) v ( p v r ) ]
~ ( p v r ) v [ (~ q ^ r ) v ( p v r ) ]
(~p ^ ~ r ) v [ (~ q ^ r ) v ( p v r ) ]
then i get stuck in here :
(~p ^ ~ r ) v (~ q ^ r ) v ( p v r )
i cant use associativity law for r because i see different logic operators
what can i do ?

Answer 1

Acutally, this simplifies dramatically to T. Consider just the first step:

(p∨r) → ((q∨¬r) → (p∨r))

And you see that this is nothing more than a special instance of the logical axiom φ → (ψ → φ), so this evaluates to T.

If you consider the way you were doing it, then at the step:

¬(p∨r) ∨ ((¬q ∧ r) ∨ (p∨r))

By commutativity:

¬(p∨r) ∨ ((p∨r) ∨ (¬q ∧ r))

By associativity:

(¬(p∨r) ∨ (p∨r)) ∨ (¬q ∧ r)

Per law of excluded middle:

T ∨ (¬q ∧ r)

Which is obviously just:

T

Answer 2

i'm stumped yet i will annoy others by making use of asserting that their solutions are incorrect because it states that there must be diverse quantities of crimson and black hats eg There could be 20 crimson and 0 black, 20 black and 0 crimson, 4 crimson and sixteen black, or the different combination, and so on. and so on. plz deliver answer

Answer 3

(~p ^ ~ r ) v (~ q ^ r ) v ( p v r ) Consider that

(~p ^ ~ r )v ( p v r ) == (p v r) v (p v r) by DeMorgan
== (p v r)

So the line you got stuck on reduces to:
(p v r) v (~q ^ r)

I think that's as simplified as you can get it.