is saying: (p→q)→(q→p) equivalent to saying p→q, q ∴ p?

Question

I'm reading a logic book and I came across some sentences of the form:

(p→q)→(q→p)
(p→q)→(~p→~q)

which can be read as:

from: if p, then q, it follows that: if q, then p;
from: if p, then q, it follows that: if not p, then not q;

my question is, since in English they sound like the arguments:

p→q
q
∴ q

and

p→q
~p
∴ ~q

do they mean the same thing?

because I read in some books very long sentences with that form, as:

p→(q→r)→(p→q→[p→r])

and I have to say that I really dont understand how I should read them, but if I could "break them down" like the arguments above I could read them as:

p→(q→r)
p→q
∴ p→r

which to me are easier to understand. Is this correct?

Answer 1

The two are not the same. The first is a sentence, and the second is an argument.

That said, there is a connection between conditionals and arguments: any conditional $\varphi \to \psi$ is a logical truth if and only if the argument:

$$\varphi$$

$$\therefore \psi$$

is logically valid.

And yes, we can break down further conditionals, e.g. $\varphi \to (\psi \to \chi)$ is a logical truth if and only if the argument:

$$\varphi$$

$$\psi$$

$$\therefore \chi$$

is logically valid. Etc.

So in your first example, we have that $(p\to q) \to (q \to p)$ is a logical truth if and only if the argument:

$$p \to q$$

$$q$$

$$\therefore p$$

is logically valid.

Note that in this particular case, the statement $(p\to q) \to (q \to p)$ is not a logical truth, and indeed its corresponding argument is not logically valid.

Of course, if we would add the statement to the premises of its correponding argument, then we can validly infer the conclusion of the original argument:

$$(p\to q) \to (q \to p)$$

$$p \to q$$

$$q$$

$$\therefore p$$

And so in that sense, you could indeed say: "aha, so given $(p\to q) \to (q \to p)$, that means that if I ever have $p \to q$ and $q$, then I can infer $p$". And I think that's very close to what you have in mind here.

Answer 2

Things like $(p\rightarrow q)$ are symbolic expressions, don't try too hard to read them in natural english. For many cases it can be tempting to always think of them in english, because expressions like $(p\wedge q)$ can be easily be translated as "$p$ and $q$". You should prefere to treat them like they are: a logic expression, characterized by the table

$$\begin{array}{cc|cc} p & q & p\wedge q & p\rightarrow q \\ \hline 1 & 1 & 1 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 1 \end{array}$$

If you need a natural language translation for $(p\rightarrow q)$, I like to think of it not like a theorem, as I think you sujest, but like a promise under condition. "If you do $p$, I will do $q$": I you do $p$ and I do $q$, I kept the promise, and if I did not do $q$ I did not keep the promise. On the other hand, if you do not do $p$, I already keept the promise whether or not I do $q$

But I reapeat, I sugest thinking of them more like abstract expressions and less like natural language expressions, that way their properties become clearer and the calculations easier. That was the whole perpose of creating abstract logic. In the end, writing complex expressions is impossible in natural language. However, as ancientmathematician mentioned, the use of parentheses is necessary to avoid ambiguity

So for expressions like $((p→q)→(q→p))$, we just follow the properties of the operation $\rightarrow$, composing the complex expression from simpler ones

$$\begin{array}{cc|cc} p & q & p\rightarrow q & q\rightarrow p & (p→q)→(q→p)\\ \hline 1 & 1 & 1 & 1 & 1\\ 1 & 0 & 0 & 1 & 1\\ 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 & 1 \end{array}$$

Notice that from the table it follows that the expression $((p→q)→(q→p))$ is equivalent to the expression $(q\rightarrow p)$. That means they have the same truth values for all possible combinations of truth values for $p$ and $q$. That is clear when we look at the table, but I would argue it is near impossible to see if you think of them in natural language.