Question

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)

Answer 1

Answer:

(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)

Answer 2

The resulting prepositions are listed below:

1. $p \,\lor\,\neg q$
2. $\neg p \,\land\,(\neg q \,\lor \,\neg p)$
3. $p\,\lor \,q$

How apply inference rules to simplify compound proposition

A proposition is an entity that contains a truth value, there are simple and compound propositions. Compond propositions are formed by simple propositions and they may be simplified by inference rules.

In this question we must use the following inference rules to simplify each compound proposition:

(1) $p \implies q \iff \neg p \,\lor\,q$   Conditional-disjunction equivalence

(2) $\neg\neg\, p \iff p$   Double negation

(3) $p \,\lor \,p \iff p$   Addition rule

(4) $\neg (p\,\land \,q) \iff \neg p\,\lor\,\neg q$   De Morgan's rule

(5) $\neg (p\,\lor \,q) \iff \neg p\,\land\,\neg q$   De Morgan's rule

(6) $p\,\land\,p \iff p$ Simplification

Now we proceed to simplify each compound proposition:

(a) $\neg p \implies \neg q$

1. $\neg p \implies \neg q$   Given
2. $\neg \neg p \,\lor\,\neg q$   Conditional-disjunction equivalence
3. $p \,\lor\,\neg q$   Double negation/Result

(b) $(p\,\lor\,q)\implies \neg p$

1. $(p\,\lor\,q)\implies \neg p$   Given
2. $\neg (p\,\lor\,q)\,\lor\,\neg p$   Conditional-disjunction equivalence
3. $(\neg p\,\land\,\neg q) \,\lor \,\neg p$   De Morgan's rule
4. $(\neg p \,\lor\,\neg p)\,\land \,(\neg q\,\lor\,\neg p )$   Distributive property
5. $\neg p \,\land\,(\neg q \,\lor \,\neg p)$   Addition rule/Result

(c) $(p\implies \neg q)\implies (\neg p \implies q)$

1. $(p\implies \neg q)\implies (\neg p \implies q)$   Given
2. $(\neg p \,\lor\,\neg q)\implies (\neg\neg p \,\lor\,q)$   Conditional-disjunction equivalence
3. $(\neg p \,\lor\,\neg q)\implies (p\,\lor \, q)$   Double negation
4. $\neg (\neg p \,\lor \,\neg q)\,\lor\,(p\,\lor\,q)$   Conditional-disjunction equivalence
5. $(\neg\neg p \,\land\,\neg\neg q)\,\lor\,(p\,\lor\,q)$   De Morgan's rule
6. $(p\,\land\,q)\,\lor\,(p\,\lor\,q)$   Double negation
7. $(p\,\lor\,(p\,\lor\,q))\,\land (q\,\lor\,(p\,\lor\,q))$   Distributive property
8. $((p\,\lor\,p)\,\lor\,q)\,\land\,((q\,\lor\,q)\,\lor\,p)$   Commutative and distributive properties
9. $(p\,\lor\,q)\,\land\,(p\,\lor\,q)$   Addition rule
10. $p\,\lor \,q$   Simplification/Result

To learn more on propositions, we kindly invite to check this verified question: https://brainly.com/question/6709166