When "if P(x), then Q(x)" is false. How to explain it with truth table?

Question

Question 1

$P(x) = x > 2$

$Q(x) = x^2 \le 4$

For all $x$ are real numbers, if $P(x)$, then $Q(x)$.

Put it simply. It is false for all $x > 2$.

But I am not sure how to explain each row in the truth table.

When $P(x)$ is true and $Q(x)$ is false.

There exists $x > 2$ and $x^2 >4$.

P(x)	Q(x)	If P(x), then Q(x)
true	false	false

But how to explain the following tables?

P(x)	Q(x)	If P(x), then Q(x)
true	true	Could I fill in this cell with the value true?

When I put the value true in the cell.

Could I just give an explanaiton such as

Definitely it would not lead to the first row.

Then in a different row.

When $P(x)$ is false and $Q(x)$ is false.

There exists $x < -2$ and $x^2 > 4$.

P(x)	Q(x)	If P(x), then Q(x)
fasle	false	Could I fill in this cell with the value true?

When I put the value true in the cell.

What explanation could I give?

Question 2

My doubts sound weird or strange.

Does it mean that I misunderstand the meaning the truth table?

Answer 1

You can use a truth table here for the truth conditions of implication $\to$, but you also need to know about the semantics of quantifiers.

---

The well-formed formula $P(x)$ on its own isn't true or false because it contains a free variable. It is only true or false when considered in a context that supplies each of the free variables (in this case just $x$) with an interpretation.

The original statement, reworded slightly, is this.

For all real numbers $x$, if $P(x)$, then $Q(x)$.

Expressed symbolically in first-order logic, it looks like this. The domain of discourse is understood to be $\mathbb{R}$.

$$ \forall x \mathop. P(x) \to Q(x) $$

A statement headed by a universal quantifier that governs a variable $x$ is false if and only if there is a value that $x$ be given that makes the body of the quantified statement false.

You have already indicated when this statement is false, namely when $x$ is greater than 2.

So let's pick a context that assigns $x$ the value $3$.

Now we have, $P(3) \to Q(3)$.

$P(3)$ is $3>2$, which is true.

$Q(3)$ is $3^2 < 4$, which is false.

$\text{true} \to \text{false}$ is false, according to the truth table of implication.

a→b
     b
     1 0
---------
a 1  1 0
  0  1 1

Since we've shown a single counterexample, the entire original "for all" statement is false.

Answer 2

We have the universal generalization $\forall x:[x\in R \to x\gt 2 \land x^2\le 4]$

You can prove it false by producing a single counter-example, $x=4$ in this case.

From the truth table for $A\to (B \land C)$

where

$~~~~A = 4 \in R~~$ (T)

$~~~~B= 4\gt 2~~$ (T)

$~~~~C = 4^2 \le 4~~$ (F)

we have $~4\in R ~\to~ 4\gt 2 ~\land~4^2\le 4~$ being false (see line 2).

Therefore, it is false that $~\forall x:[x\in R \to x\gt 2 \land x^2\le 4]$