What we can infer from there exists x satisfying P

Question

If it is known that there exists x satisfying P, can we infer that there also exists x not satisfying P?

I ask this question since I have a problem as follows.

Given three premises: (1)if a student likes math, he doesn't like social studies. (2)every student likes math or social studies. (3)some students like math.

can we come to a conclusion that there exists student who doesn't like math?

Answer 1

NO.

Assume a domain with only the number $0$. We have that $P(x) := (x=0)$ is satisfied while $\lnot (x=0)$ is not.

We say that we can infer or deduce $\psi$ from $\varphi$ when $\psi$ is a logical consequence of $\varphi$; in symbol : $\varphi \vDash \psi$.

It can be useful to go back to the source of formal logic : Aristotle for the notion of valid argument, i.e. of an argument that is "justified by virtue of form alone".

In Aristotle's logic :

A deduction is speech (logos) in which, certain things having been supposed, something different from those supposed results of necessity because of their being so [emphasis added]. (Prior Analytics I.2, 24b18-20)

The core of this definition is the notion of “resulting of necessity” . This corresponds to a modern notion of logical consequence: X results of necessity from Y and Z if it would be impossible for X to be false when Y and Z are true. We could therefore take this to be a general definition of “valid argument”.

Aristotle proves invalidity by constructing counterexamples. This is very much in the spirit of modern logical theory: all that it takes to show that a certain form is invalid is a single instance of that form with true premises and a false conclusion. However, Aristotle states his results not by saying that certain premise-conclusion combinations are invalid but by saying that certain premise pairs do not “syllogize”: that is, that, given the pair in question, examples can be constructed in which premises of that form are true and a conclusion of any of the four possible forms is false.

Thus, my counterexample above shows that :

$P(x) \nvDash \lnot P(x)$

i.e. from the fact that there exists $x$ satisfying $P$, we cannot infer (or deduce) that there also exists $x$ not satisfying $P$ (i.e. satisfying $\lnot P$).