What does implication actually mean?

Question

I recently have been reading a book on proofs, and in the very first chapter it started to discuss implication. It gave implication's truth table and different ways of expressing implication, and it did attempt to explain it to the reader, but I still found myself confused on what implication actually is.

Implication's truth table

Implication with words

Explanation of Implication

My confusion lies in what people say implication logically represents. Based on how its expressed with words "If A then B" you would think it meant "If A is true then B must be true". However, you have cases where something called vacuose truth applies(Another part of implication that I don't understand), which contradicts the initial assumption about implication with statements like "False ⇒ True" being true.

I've seen videos where a college proffesor said it shows "the logical relationship between A and B" which didn't make sense to me, and forum posts from this site and others trying to explain it but each time I read them I find myself as or even more confused.

I'm sorry if I sound frustrated or angry. I just really want to understand proofs, but I've found this one concept to be a roadblock to understanding the rest of the book. I'm also sorry if the format of the post is a little weird. This is the first time ever posting on this site, and I'm not sure how to do the fancy math format that I usually see on Math stackexchange. Any help trying to get me to understand this concept would be greatly appriciated!

Answer 1

You can think of $A \to B$ as asking "If I give you $A$ can you return $B$?".

This line of thinking is based on a more general logic than classical logic, but here you don't need to know much about it except

• "$P$ is true" should be understood as "we have $P$"
• "$P$ is false" should be understood as "there is no way we can get $P$"

So with this in mind let's look again at the truth table and what the answer is in every case.

• No matter whether $A$ is true or false, if $B$ is true i.e. we have $B$, then we can definitely return $B$. So we have $A \to B$.
• If $A$ is true and $B$ is false, i.e. there is no way to get $B$, then there is no way we can get $A \to B$.
• The last case might still seem a bit strange: If $A$ is false and $B$ is false, i.e. there is no way to get $A$, then the answer to "If I give you $A$ can you return $B$?" is "Well apparently then everything is possible, so I can also get you $B$".

(In the end implication is -in this view- connected to functions. I wrote more about this here)

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It's worth thinking about the "$A$ is false implies everything" a bit more.

Let's say we restrict our interest to statements involving natural numbers only (very concretely I mean the first-order theory of Peano arithmetic) and assume we don't have -as of yet- any idea what a false statement is meant to be.

There is however a prototypical very problematic statement; namely $0 = 1$.

How problematic? Very much so! Because combined with induction, $0 = 1$ can be used to show $x = 0$ for every term $x$ and this enables us to show $0 = 1 \to B$ for every statement $B$. (every equation just becomes trivial if everything is $0$)

We can now go ahead and define a statement $A$ to be false iff $A \to 0=1$. With this definition then, do we have $A \to B$ when $A$ is false? Well in this case

• We are given $A$, we know $A$ is false and we want to return $B$.
• $A$ false means $A \to 0=1$ and since we are also given $A$, we get $0=1$.
• From $0 = 1$ we can get anything so in particular $B$.

So you can think of false statements concretely as something like $0 = 1$, which really does make it possible to show anything in Peano arithmetic.

Answer 2

In a proof, an implication must be taken in its formal meaning and has little to do with an implied "cause-effect" relation. It is implicit that when you write an implication $P\implies Q$, the result of the evaluation of this boolean expression is true.

The proposition

$$n>0\implies n+1>0$$

can be true in three ways and false in one, namely

• $n>0\land n+1>0,$
• $\color{red}{n>0\land n+1\le0},$
• $n\le0\land n+1>0,$
• $n\le0\land n+1\le0.$

As you can check explicitly, the red proposition is impossible while all others are possible, hence the implication always has a true value. Also note that the reverse relation is not an implication,

$$n+1>0\ \not\!\!\!\implies n>0$$

because of the counterexample

$$\color{red}{0+1>0\land 0>0}$$ which is yields a false value.