p,q are two propositions.It is given that, p ⇒ q is true.Consider the following conclusions,

Question

1. $ \neg p\rightarrow\neg q$ is true
2. $\neg q\rightarrow\neg p$ is true
3. $p\rightarrow \neg p∨q$ is true

Now which one is the correct? and explain this.Thanks!

Answer 1

NOTE: $\sim p$ means "not p", $\sim q$ means "not q".

1. $\sim p \implies \sim q$ is called the inverse of a conditional statement. The inverse is not always true given the original.

For example, if $p$ is false and $q$ is true, then we know the original is true because it has a false hypothesis, but the inverse is false because it has a true hypothesis, but a false conclusion.

Therefore, we can not conclude this.

2. $\sim q \implies \sim p$ is called the contrapositive of a conditional statement. The contrapositive is always true given the original, so we can conlude that this is true.
3. $p \implies (\sim p \vee q)$ is something I've never seen before, but it means we have to assume $p$ and then prove $\sim p \vee q$.

Thus, let's assume $p$.

We know that $p \implies q$, so with $p$, we can conclude $q$.

Now, since we have $q$, we also have $v \vee q$, where $v$ can be absolutely anything, since anything in a conjunction with a true statement is true.

Therefore, we can make $v=\sim p$ to conclude that $\sim p \vee q$.

Thus, we have successfully proven $\sim p \vee q$ given $p$, so we can conclude $p \implies (\sim p \vee q)$.

If you want to check my reasoning, you can do what @ncmathsadist did and make a truth table for all of these statements.