Propositional Logic Question Related to Understanding "→" and Tautologies

Question

I should preface this with that I have never studied logic before. When answering my question, please assume that I know nothing about formal logic.

Just now, I was reading a different question and one of answers gave the statement:

$(A⇒B)⇔(¬B⇒¬A)$

(Colloquially known as: the statements "Paris is in France" and "Not being in France means not being in Paris" mutually imply each other.)

Initially, I got confused and understood $(A⇒B)⇒(¬B⇒¬A)$, but not $(¬B⇒¬A)⇒(A⇒B)$.

Then I realized I was mistakenly thinking of the "$⇒$" in $(¬B⇒¬A)$ as an "$∧$", in which case $(¬B∧¬A)⇏(A⇒B)$.

However, this simple issue raised another more fundamental question in my mind:

What is propositional logic assuming in such a statement for $(¬Q⇒¬P)⇒(P⇒Q)$ to always be True?

Can't there be a case, where $¬P$ is a tautology, in other words, $(¬Q⇒¬P)∧(Q⇒¬P)$, and thus in this case, $(¬Q⇒¬P)⇏(P⇒Q)$?

This would be assuming that, in a case where $⊨P$ ("$P$ is a tautology"), that $Q⇒P$ is True. In this case, isn't the above a contradiction? What I'm I confusing here?

Conversely, if I assume that in a case where, $⊨P$, that $Q⇒P$ is False, this would seem to suggest that "$⇒$" would denote more than simply the truth value of a proposition. Since, $Q⇒P$ would be False even though $P$ is always True when $Q$ is True, which I now understand would be blatantly wrong.

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(My question can be boiled down to simply not distinctly understanding what the $→$ logical operator does, i.e. in terms of its truth table where, only the proposition $True→False$ is False, and the propositions, $True→True$, $False→True$, and $False→False$, are all True. This is evidently a common confusion among laymen and new students alike, since it's non-intuitive to consider what "if... then..." (or even worse, "implies") means in the context of $False→True$ and $False→False$ propositions (i.e. propositions with a False antecedent), when a proposition can only be evaluated as True or False.)

Answer 1

" Can't there be a case, where $¬P$ is a tautology, in other words, $(¬Q⇒¬P)∧(Q⇒¬P)$, and thus in this case, $(¬Q⇒¬P)⇏(P⇒Q)$? "

In propositional logic, a tautology is a proposition that is true by virtue of its truth-functional form. As such, $¬P$ is patently not a tautology, merely that it is (being interpreted as) true, i.e., that it is a true statement. [In other words, we distinguish the claim “$\lnot P$ is always true (whichever the truth value of $Q$)” from the claim “$\lnot P$ is always true (in truth-functional form)”.]

So, since $¬P$ is true (and $P$ thus false), the meta-statement $$(¬Q\rightarrow ¬P)⇏(P\rightarrow Q)\tag1$$ has both a true antecedent and a true consequent (since both $x\rightarrow T\;$ and $\;F\rightarrow x$ are by definition true). As such, and since $T\rightarrow T$ is by definition true, it must be that $$(¬Q\rightarrow ¬P)⇒(P\rightarrow Q),$$ i.e., statement (1) is false.

Answer 2

In propositional logic, if $Q$ is a theorem, then so is $P\to Q$, for any choice of $P$. It helps to think of "$P\to Q$" as saying "knowing that $P$ is true is sufficient to know that $Q$ is true". It's not saying anything about whether $P$ "causes" $Q$, it's just describing the pairs of truth values that we consider possible. We think of each proposition as a fully-qualified sentence about the world, so no proposition is fundamentally "conditionally true" or true "because of" another proposition; each one is just either a true statement about the world or a false one.

First-order logic (propositional logic with variables and quantifiers) is a little more expressive: you can have $P(t)\to Q(t)$ be true for a particular term $t$ while the "for all" sentence $\forall x(P(x)\to Q(x))$ is false. In this case, you might think of the former sentence as an "accidental" implication and the latter as stating more of a "real" one. (But note that $\forall x(P(x)\to Q(x))$ is derivable from $\forall xQ(x)$, so it can still be "accidentally true".)

You might also be interested in modal logic, which formalizes the notions of "necessary" and "possible" propositions.

Answer 3

I'm guessing you are too much biased for 'imply' in a natural language to interpret $\Rightarrow$.

In propositional logic, $P \Rightarrow Q$ is false only when $P$ is true and $Q$ is false. And as noted in your question $P\Rightarrow Q$ is true if $Q$ is true. This is the definition.

So for example,

• 'Berlin is in France' $\Rightarrow$ 'Paris is in France'
• $1 = 0$ $\Rightarrow$ $2+2=4$

are both true in terms of propositional logic because Paris is in France and $2+2=4$, however weird these sound.

Regarding your question, $(Q\Rightarrow \neg P) \land (\neg Q \Rightarrow \neg P)$ can be true if $\neg P$ is a tautology, as you say. But that does not entails $(\neg Q\Rightarrow \neg P)\not\Rightarrow (P\Rightarrow Q)$.

In order to 'prove' that $(\neg Q\Rightarrow \neg P)\not\Rightarrow (P\Rightarrow Q)$ can happen, you need to find a case where $\neg Q \Rightarrow \neg P$ is true and $P\Rightarrow Q$ is false. (Here $\neg Q \Rightarrow \neg P$ replaces $P$ and $P\Rightarrow Q$ replaces $Q$ in the first sentence of the 2nd paragraph above, the definition of $\Rightarrow$).

But this cannot happen: $P\Rightarrow Q$ is false means that $P$ is true and $Q$ is false, that is $\neg Q$ is true.Then, since $\neg Q\Rightarrow \neg P$ is true, $\neg P$ is also true. Then we have both $P$ and $\neg P$ true, a contradiction.

I hope this helps you understand where you are confused, but probably you should check any standard textbook first. It will explain the same thing as above more properly. (I find the words above too clumsy...)

Answer 4

The formal way to prove this kinds of statements , is to compare for each A,B, that the result is the same. You can easily verify that these two are just the same.