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The “implies” connective “$\rightarrow$” is one of the stranger connectives in propositional logic. Below are a series of statements regarding implications. Which of the following statements is/are TRUE?

- For any propositions $P$ and $Q,$ the following is always true: $(P \rightarrow Q) \vee (Q \rightarrow P).$
- For any propositions $P, Q$ and $R$, the following statement is always true: $(P\rightarrow Q) \vee (Q \rightarrow R).$
- For any propositions $P, Q$ and $R$, the following statement is always true: $(P\rightarrow Q) \vee (\neg P \rightarrow R).$
- For any propositions $P, Q$ and $R,$ the following statement is always true: $(P \rightarrow Q) \vee (R \rightarrow Q)$.

1 vote

$A$. For any propositions P and Q, the following is always true: $(P \rightarrow Q) \vee (Q \rightarrow P).$

$\text{Proof}$:

Here's one way to see this. If $Q$ is true, then $P \rightarrow Q$ is true because anything implies a true statement. If $Q$ is false, then $Q \rightarrow P$ is true because false implies anything. (If this is confusing, you should review the truth table for $\rightarrow !$)

$B$. For any propositions $P, Q$, and $R$, the following statement is always true: $(P \rightarrow Q) \vee (Q \rightarrow R).$

$\text{Proof}$ :

This is basically the same argument as before. If $Q$ is true, then $P \rightarrow Q$

is true because anything implies a true statement. If $Q$ is false, then $Q \rightarrow R$ is true because false implies anything.

Of all the connectives we've seen, the $\rightarrow$ connective is probably the trickiest. We asked this question to force you to disentangle notions of correlation or causality from the behavior of the $\rightarrow$ connective.

$\text{Proof}$:

Here's one way to see this. If $Q$ is true, then $P \rightarrow Q$ is true because anything implies a true statement. If $Q$ is false, then $Q \rightarrow P$ is true because false implies anything. (If this is confusing, you should review the truth table for $\rightarrow !$)

$B$. For any propositions $P, Q$, and $R$, the following statement is always true: $(P \rightarrow Q) \vee (Q \rightarrow R).$

$\text{Proof}$ :

This is basically the same argument as before. If $Q$ is true, then $P \rightarrow Q$

is true because anything implies a true statement. If $Q$ is false, then $Q \rightarrow R$ is true because false implies anything.

Of all the connectives we've seen, the $\rightarrow$ connective is probably the trickiest. We asked this question to force you to disentangle notions of correlation or causality from the behavior of the $\rightarrow$ connective.

0 votes

We know that p → q is equivalent to ~p v q. and similarly ~p v p is always true .Using these simple equivalence rule.

We can solve the problem.:

For option 1,2 and 3:

(p → q) v (q → p) can be written as (~p v q) v (~q v p) and we know ~p v p is true.

Hence option 1 is true similarly option 2 and 3.

But for last option this is not valid ~p v q v ~r V q.Hence it is contingency.

We can solve the problem.:

For option 1,2 and 3:

(p → q) v (q → p) can be written as (~p v q) v (~q v p) and we know ~p v p is true.

Hence option 1 is true similarly option 2 and 3.

But for last option this is not valid ~p v q v ~r V q.Hence it is contingency.