True/false? The negation of "If it snows, then the train isn't on time." is "When the train is on time, then it doesn't snow."

Question

True/false? The negation of "If it snows, then the train isn't on time." is "When the train is on time, then it doesn't snow."

This is a confusing task, I tried to give these words variables.

So let

$$a= \text{snow}$$

$$b = \text{not on time}$$

First sentence: $a \rightarrow b$

Second sentence: $\neg b \rightarrow \neg a$

Thus the statement is false beccause $a \rightarrow b \neq b \rightarrow a $

But I'm really not sure on that : /

Answer 1

You're reaching a correct conclusion, but saying "because $a\to b\neq b\to a$" doesn't work as an argument for it.

To begin with, you're just asserting this non-equality, not arguing for it -- and at the level of this exercise I think you're supposed to provide some kind of argument, such as a truth assignment for $a$ and $b$ that gives the two sides different truth values.

More seriously, it is not clear at all what $b\to a$ has to do with anything in your argument. I'm getting afraid that you think the negation of $a\to b$ should be $b\to a$ or something like that, which is not the case.

The negation of $a\to b$ is $a\land \neg b$. So negating "if it snows then the train is not on time", should be, "it snows, and yet the train is on time".

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A more subtle thing here is that one should probably not represent the English "if -- then" as merely $\to$ in the original sentence. Representing it as just $a\to b$ implicitly makes it into a claim about one particular time (or one particular possible world), whereas the English sentence seems to express a more general judgement: "At every relevant time when it snows, it will also hold that the train is not on time". (Let's leave the ambiguity about which times are "relevant" unresolved; that's not the interesting point).

Negating the sentence above would also invert the implied quantifier, so we should end up with something like

There is (was, may be, will be, could be?) an occasion when it snows and yet the train is on time.

Unfortunately, the exercise is probably meant to be one of propositional logic where this cannot even be clearly expressed, and authors/teachers are generally not much respectful of how it violates the meaning of natural-language sentences to shoehorn them into the propositional mold.

Answer 2

Actually, the two statements are equivalent, rather than the negation of each other! So, the statement is False.

Your symbolizations are correct, though it is better to use "the train is on time" for $b$ rather than the train is not on time ... when you have a 'not', it is better to treat that as a logical operator. So you get:

$a \rightarrow \neg b$

and

$b \rightarrow \neg a$

And they are equivalent, since they are each other's contrapositive. That is, when you have a conditional, you can swap the two sides, and negate both sides, and the result will be equivalent: $P \rightarrow Q \Leftrightarrow \neg Q \rightarrow \neg P$

To see this, take a simpler example:

'if it rains, the streets are wet'

Given this, what do you know if the streets are not wet? You know that it doesn't rain, of course! So from the above statement, you can infer:

'if the streets are not wet, it does not rain'

And it turns out you can go the other way around as well: if you know that 'if the streets are not wet, it does not rain', and it does rain? Then the streets must be wet, since if they are not, it shouldn't be raining. So, we can go back to 'if it rains, the streets are wet'.

In other words, these two sentences are logically equivalent.

In your case, the two sentences given to you are equivalent as well by this very 'contrapositive' principle. So no, the second one is not the negation of the first.

Answer 3

Short Answer: The highlighted statement at the top of your post is false.

Long answer:

You've defined your propositions to be

(a) It snows.

(b) The train is not on time.

In $(1)$ I'll list the statement you are to negate, and its contrapositive, which is equivalent to it: $$a \to b \equiv \lnot b \to \lnot a\tag 1$$

So the negation of $a \to b$ is not given by $\lnot b \to \lnot a$. In fact, as I mention above, the two statements are logically equivalent.

Nor is the negation of $a\to b$ given by $\lnot a \to \lnot b \equiv b\to a$, which is the inverse of $a\to b$.

The negation of any statement $p$ is simply $\lnot p$. In this case $p=: a\to b$.

So the negation of $(a \to b)$ is given by $\lnot(a \to b)$:

$$\begin{align} \lnot(a \to b) &\equiv \lnot(\lnot a \lor b) \tag{implication}\\ \\ & \equiv a \land \lnot b\tag{Demorgan's}\end{align}$$