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

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?

• 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. Jan 9 at 0:05
• 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 $Q(x).$" Jan 9 at 0:09
• That is the correct negation of "For all $x,$ if $P(x)$ then $Q(x),$" yes. @StatsCruncher Jan 9 at 0:26
• But the key point is that truth tables are meant for propositional logic, which is logic without "for all" and "there exists." Jan 9 at 0:27
• @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. Jan 9 at 6:04

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.

• Is that table or diagram a formal truth table? Why is it quite different? Jan 9 at 8:26
• 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. Jan 9 at 19:06
• Now I get It right. Appreiciate your further explanation. 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 $$A\to (B \land C)$$

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 $$~\forall x:[x\in R \to x\gt 2 \land x^2\le 4]$$