Does an implication have to be true in all interpretations to be true?

110 Views Asked by At

I am wondering about this example:

$P: x = 3$

$Q: x^2 = 9$

To me it seems clear that $P \implies Q$ is true. If $x=3$ it is true that $x^2=9$.

But what about the converse here?

The statement $Q \implies P$ "feels" false to me at first, but is it? If $x^2=9$ is it true that $x=3$? It could be $-3$, but it's possible that $x=3$ as well, so it is possibly true, which means it is not always false. This statement is sometimes true, sometimes false.

So how do we evaluate it, and what rule in logic tells us how to interpret it?

1

There are 1 best solutions below

0
On BEST ANSWER

First, there is a distinction between true/false and valid/invalid. True/false are usually used with respect to a given model. Valid(which corresponds to provable for typical logics) means "true in all models".

We don't usually talk about open formulas, i.e. formulas containing free variables like $x$ in your example, as being true/false or valid/invalid. Usually, only closed formulas with no free variables are allowed to be considered to be true/false or valid/invalid.

One common convention is to treat all free variables as implicitly universally quantified when asking about their truth/validity. In your example, this would mean considering $\forall x.Q\implies P$ which is false in e.g. a model of the reals where the operations are interpreted as usual, but true in a model of the naturals as dxiv mentions. This assumes those are models which depends on the full theory you are using which you haven't specified. As such, whether the statement is valid or not depends on the full theory, though in $\mathbb R$ with the usual interpretation of the operations is a model, then it is not valid.

Alternatively, you can consider an open formula as specifying a relation on a given domain set. In your example, taking the domain set to be $\mathbb R$ (assuming it's a model), the formula $Q\implies P$ would be $\{x\in\mathbb R\mid x^2=9 \implies x=3\}=\mathbb R -\{-3\}$ because the formula is true for all $x\in\mathbb R$ except for $x=-3$. Connecting to the previous paragraph, $\forall x.\varphi(x)$ is true (for a given domain, $D$) when $\{x\in D\mid \varphi(x)\}=D$. Thus $\mathbb R-\{-3\}\neq\mathbb R$ is another way of saying the universally quantified formula is false (in this model, but if it is false in any model then it is invalid). If we use $\mathbb N$ as the domain set (again, assuming this is a model of your theory), then $\{x\in\mathbb N\mid x^2=9\implies x=3\}=\mathbb N$ corresponding to the universally quantified formula being true in this model.

To reiterate though, the interpretation of a formula with $n$ free variables is a $n$-ary relation on the domain set. A closed formula corresponds to $n=0$, and a $0$-ary relation is just a subset of a singleton set and is thus either the singleton set itself or empty, corresponding to true and false respectively.