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?

  • 1
    $\begingroup$ It's an implication. You assumed $P(x)$ was true and saw $Q(x)$ was false. Then the implication is false (this is just a rule). Using a truth table isn't even really helpful here. You really just need to know when $P \Rightarrow Q$ is true. $\endgroup$
    – davinci_07
    Jan 9 at 0:05
  • 1
    $\begingroup$ The statement "If $P(x)$ then $Q(x)$" is very different from "For all $x,$ if $P(x),$ then $Q(x).$" The latter cannot be expressed with truth tables. Your example is a particular strong example, where it is true that: "For all $x,$ if $P(x)$ then not $Q(x).$" That is very different from "It is not true that for all $x,$ if $P(x)$ then $Q(x).$" $\endgroup$ Jan 9 at 0:09
  • 1
    $\begingroup$ That is the correct negation of "For all $x,$ if $P(x)$ then $Q(x),$" yes. @StatsCruncher $\endgroup$ Jan 9 at 0:26
  • 2
    $\begingroup$ But the key point is that truth tables are meant for propositional logic, which is logic without "for all" and "there exists." $\endgroup$ Jan 9 at 0:27
  • 1
    $\begingroup$ @StatsCruncher Please try to use Mathjax to typeset questions in the future. I edited this question to use Mathjax to help get you started. The syntax is very similar to LaTeX, if you're familiar with LaTeX. $\endgroup$ Jan 9 at 6:04

2 Answers 2


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.

     1 0
a 1  1 0
  0  1 1

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

  • $\begingroup$ Is that table or diagram a formal truth table? Why is it quite different? $\endgroup$ Jan 9 at 8:26
  • 1
    $\begingroup$ It’s a truth table for the connective $\to$. I’m just organizing it as a 2x2 table rather than as a table with four rows with three columns. $\endgroup$ Jan 9 at 19:06
  • $\begingroup$ Now I get It right. Appreiciate your further explanation. $\endgroup$ Jan 9 at 21:18

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 $A\to (B \land C)$


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

enter image description here

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


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