Why isn't the negation of "p implies q" "p implies not q"?

Question

My line of reasoning is that if it is not possible for both statements to be true at the same time and both statements to be false at the same time, then the two statements are negations of each other. Clearly, it is not possible for "p implies q" and "p implies not q" to be true at the same time since if p is true then q would have to be both true and false. To prove that "p implies q" and "p implies not q" cannot be false at the same time, assume for the sake of contradiction that both are false. Then "p and not q" and "p and q" must both be true since the negation of "p implies q" is "p and not q". But this is impossible. Thus "p implies q" and "p implies not q" cannot be false at the same time, and "p implies q" and "p implies not q" are negations of each other.

Answer 1

I don't have the time to read your wall of text, so let me make my point briefly.

If I claim $P \Rightarrow Q$ and I'm wrong, how could that be? That should be evident when $P$ holds and $Q$ doesn't, and nothing else really "shows" it's false that $P$ implies $Q$. That is the motivation for wanting $\sim(P \Rightarrow Q)$ to be $P \& \sim Q$. So we define the truth values of $P \Rightarrow Q$ to make that work.

Answer 2

$P$ implies $Q$ can be translated into

1. If $P$ is the case, then necessarily $Q$ is also the case.

That statement is false if there is a case when I have $P$ but I do not have $Q$. That doesn't mean not having $Q$ will always follow from $P$; all I need is one instance of that happening and proposition $1$ will be false. This makes sense, because $P \implies Q$ means I will always have $Q$ if $P$ is given, so to say the opposite is to say $P$ may ocurr (not must occur) without $Q$.

For example, the proposition

2. If it rains, then I get wet

is false because there are times when it rains and I do not get wet (I stay home, let's say). So to say it is false is to say

3. There are times when it rains and I do not get wet

not

4. If it rains, I do not get wet

Notice that $3$, the correct negation of $2$, is not an implication. I am not saying rain implies me not getting wet. I'm just saying that may sometimes occur. So it doesn't take an implication to negate an implication, but only a single case (or more) of the antecedent $P$ ocurrying without the consequent $Q$.

Answer 3

To be negations, the statements must have opposing truth values for all assignments of the literals.

They do not.

$$\begin{array}{c:c|c:c}p&q&p\to q&p\to\neg q\\\hline \top&\top&\top&\bot&\checkmark\\\top&\bot&\bot&\top&\checkmark\\\bot&\top&\top&\top&\star\\\bot&\bot &\top&\top&\star\end{array}$$

However, $p\to q$ and $p\wedge\neg q$ do.

$$\begin{array}{c:c|c:c}p&q&p\to q&p\wedge\neg q\\\hline \top&\top&\top&\bot&\checkmark\\\top&\bot&\bot&\top&\checkmark\\\bot&\top&\top&\bot&\checkmark\\\bot&\bot &\top&\bot&\checkmark\end{array}$$

Answer 4

When we say that $p\implies q$ we mean that $q$ is true whenever $p$ is true. For this to be false, we only require one instance where $p$ is true and $q$ is false. But if we say that $p\implies\lnot q$ we mean that $q$ is false for every instance that $p$ is true.

For example, let $p$ be the statement "is a person" and let $q$ be the statement "is diabetic". Then $p\implies q$ says that all persons are diabetic. To prove that is false, we only need to demonstrate that one person is not diabetic. It certainly is not true that all persons are not diabetic which is what $p\implies\lnot q$ says.

Answer 5

So the reason why my proof is false is because the statement "It is not possible for "p implies q" and "p implies not q" to be true at the same time" is false. It's false because if p is false, then both "p implies q" and "p implies not q" are true. Credit to @NoahSchweber

Answer 6

It's not unreasonable to object to the classical account of negated conditionals. Classical logic is a product of the 19th century, and logicians of prior eras gave accounts of negated conditionals that differ from the classical one. In particular, Aristotle and Boethius seem to have endorsed the following propositions as logically true (at least in some contexts):

A1: $\neg (\neg A \to A)$
A2: $\neg (A \to \neg A)$
B1: $(A \to B) \to \neg (A \to \neg B)$
B2: $(A \to \neg B) \to \neg (A \to B)$

A non-trivial logic which counts these formulas, but not $(A \to B) \to (B \to A)$, among its theorems is called "connexive". Storrs McCall and R. B. Angell initiated modern study of connexive logics in the 1960s and it remains an active area of research. It turns out to be challenging to construct a connexive logic with intuitive semantics and without too many "weird" properties.

• Connexive Logic - Stanford Encyclopedia of Philosophy
• Connexive Logics: An Overview and Current Trends (Omori & Wansing, 2019)