For each of these compound propositions, use the conditional-disjunction equivalence (example 3) to find an equivalent compound proposition that does not involve conditionals.
(a) ~p→ ~q
(b) (p v q) → ~p
(c) (p→ ~q) → (~p →q)

(a) p v ~q

(b) ~(p v q) v ~p

(c) ~(~p v ~q) v (p v q)

Step-by-step explanation:

The conditional-disjunction equivalence is:

P→Q ⇔ ~P v Q

To find an equivalent compound proposition without the conditionals (without the "→") you have to apply the previous equivalence and simplify if possible.

a) ~p→~q

In this case, P= ~p and Q= ~q

Applying the equivalence:

~(~p) v ~q

p v ~q

b) (p v q) → ~p

In this case P = (p v q) and Q= (~p)

Applying the equivalence:

~(p v q) v ~p

c) (p→~q) → (~p→q)

In this case, you have to apply the conditional-disjunction equivalence for every conditional in the compound proposition.

First, let P= (p→~q) and Q= (~p→q)

~ (p→~q) v (~p→q)      (1)

Now, you have to find an equivalent compound proposition for both (p→~q) and (~p→q)

For (p→~q):

Let P= p and Q=~q

~p v ~q    

For (~p→q)

Let P= ~p and Q= q

~(~p) v q

p v q    

Then the expression (1) is:

~(~p v ~q) v (p v q)

relating to or denoting a system of numbers and arithmetic based on the number ten, tenth parts, and powers of ten.

step-by-step explanation: