How to explain this in propositional logic?

Question

Suppose I am given the statement $a \rightarrow b$ (1), and I know $b$ is true. Can I also conclude $a$ is true from just the given information (1)? My logic for trying to figure out this was — Let me apply the ¬ operator to the given statement (1). Then, $¬ a \rightarrow ¬ b$ (2). ( I am not sure if this can be done.) Then, if it were the case that $a$ is false, then $¬ a$ would be true. Which would also imply from (2) that $¬ b$ is true, $\implies b$ is false. This is a contradiction to the information that we are given that $b$ is true. So, $a$ must be true.

What confuses me is that for the general truth table of $a \rightarrow b$, there is a particular case where $a=0, b=1$ and $a \rightarrow b$ is also $1$.

Edit: My confusion arose after seeing this being done in class. I had a statement I) $p \rightarrow q$ and II) $¬ q$. II) is true. From these two, the instructor concluded that $¬ p$ is true. I don’t quite understand how that happened.

Answer 1

No. This is known as the fallacy of affirming the consequent.

Applying negation to the implication yields $\neg(a \to b)$, not $\neg a \to \neg b$, and they are not equivalent.

I also don't see where you're trying to get by applying negation to the implication. If what you want to do is a proof by contradiction, you would have to assume that $a$ is false, then in combination with the other assumptions derive a contradiction from that assumption and conclude that $a$ must be true after all.

But there is no contradiction: $a \to b$ can be true when just $b$ is true and $a$ is false. There is nothing more to it than the fact that that's what the truth table states.

If you find this unintuitive, try to avoid translating "$\to$" in your mind as an English "if $\ldots$ then", and think of it as "(not $a$) or ($b$)" or "the only thing that must not happen is for $a$ to be true but $b$ false" instead, since that's the truth conditions $\to$ is intended to express.

There are many explanations of why these are the truth conditions; see for example In classical logic, why is $p \to q$ True if $p$ is False and $q$ is True?.

Re. your edit: It's different when $b$ is false. According to the truth table of implication, in order for $a \to b$ to be true it must be the case that either $a$ is false or $b$ is true, or in other words, what can not be the case is that $a$ is true but $b$ is false. So if $a \to b$ is true but $b$ is false, then $a$ must be false as well, since otherwise you'd get the line in the truth table in which $a \to b$ is false. In contrast, there are two possible lines in the truth table of $a \to b$ and $b$ being true, one where $a$ is true and one where $a$ is false, so you can not definitely conclude the truth of $a$ just from the truth of $a \to b$ and $b$.

Answer 2

You're making some errors in your reasoning.

(i) From $a \Rightarrow b$, if b is true we cannot conclude that a is true. For example, if f is differentiable then f is continuous; but if f is continuous f doesn't need to be differentiable.

(ii) Let say $a \Rightarrow b$. Then, the implication is false only in the case that "a" is true but "b" is false. The case you say where $a=0,b=1$ corresponds to the case when the hypothesis is false, hence there's no implication so the statement is true. It might seem strange but try to reason why $\emptyset \subset A, \forall A$.

(iii) Finally, the negation of a statement is done as it follows: $a \Rightarrow b$ is equivalent to $\lnot b \Rightarrow \lnot a$. For example, the following definitions of inyective functions is equivalent: $f(x_1)=f(x_2) \Rightarrow x_1=x_2$ or $x_1 \not = x_2 \Rightarrow f(x_1) \not = f(x_2)$.