# Equations and truth tables

Let's say I work on the inequality $$|x-1|>x-2$$ in the following way:

$$|x-1|>x-2 ⇔ x-1≥0 \quad ∧ \quad |x-1|>x-2 \quad ∨ \quad x-1<0 \quad ∧ \quad |x-1|>x-2 \\ ⇔ x-1≥0 \quad ∧ \quad x-1>x-2 \quad ∨ \quad x-1<0 \quad ∧ \quad -(x-1)>x-2 \\ ⇔ x-1≥0 \quad ∧ \quad -1>-2 \quad ∨ \quad x-1<0 \quad ∧ \quad -(x-1)>x-2$$

In $$x-1≥0 ∧ -1>-2$$ there is the true statement $$-1>-2$$ . So the truth value of the whole statment $$x-1≥0 ∧ -1>-2$$ only depends on $$x-1≥0$$ . If $$x-1≥0$$ is true, then the whole statment is true, if $$x-1≥0$$ is false then whole statment is false. So I should be able to write the following $$x-1≥0 ∧ -1>-2 ⇔ x-1≥0$$ I think that I can prove that I am allowed to write $$x-1≥0 ∧ -1>-2 ⇔ x-1≥0$$ with the help of a truth table in the following way: We know that the statement $$B$$ is true. With this knowledge we analyse the statement $$A∧B⇔A$$ with the help of a truth table:

$$A$$ $$B$$ $$A$$ $$∧$$ $$B$$ $$⇔$$ $$A$$
1 1 1 1 1 1 1
0 1 0 0 1 1 0

1 for "true", 0 for "false". The first two columns show all the possible truth values the statements A and B can have.

As you can see in the truth table, the equivalence ⇔ has always the truth value of 1. This proves that if you have a statement B that is true, you can write $$A∧B⇔A .$$ So coming back to the equations: this proves that I am allowed to write $$x-1≥0 ∧ -1>-2 ⇔ x-1≥0 .$$

I also analysed this $$A∨B⇔A$$ statement in a similar way: We know that the statement B is false. With this knowledge we analyse the statement $$A∨B⇔A$$ with the help of a truth table:

$$A$$ $$B$$ $$A$$ $$∨$$ $$B$$ $$⇔$$ $$A$$
1 0 1 1 0 1 1
0 0 0 0 0 1 0

As you can see in the truth table, the equivalence ⇔ has always the truth value of 1.

So first I want to know, up to this point, is everything correct?

I have another question which is similar. Let's say I work on a set of equations and get to the following point: $$\frac{20}{x} = \sqrt{(41-x^2 )} ∧ y=\frac{20}{x}$$ Now, in a side-calculation, I work on the left equation by squaring it $$\frac{20}{x} = \sqrt{(41-x^2 )} ⟹ \frac{400}{x^2} =41-x^2$$ Now I use this Information in my set of equations, like this: $$\frac{20}{x} = \sqrt{(41-x^2 )} ∧ y=\frac{20}{x} ⟹ \frac{400}{x^2} =41-x^2 ∧ y=\frac{20}{x}$$

I want to check if I actually can write the Implication-Arrow "⟹" there. I do this by using a truth table again, in the following way: We know that $$A⇒C$$ is true (thats the squaring in the side-calculation). With this knowledge we analyse the statement $$A ∧ B ⇒ C ∧ B$$ with the help of a truth table:

$$A$$ $$B$$ $$A$$ $$∧$$ $$B$$ $$⟹$$ $$C$$ $$∧$$ $$B$$
0 1 0 0 1 1 ? ? 1
0 0 0 0 0 1 ? 0 0
1 1 1 1 1 1 1 1 1
1 0 1 0 0 1 1 0 0

I put the question marks there because if $$A$$ is false, we don't know what truth value $$C$$ has, even if $$A⇒C$$ is a tautology. All we know from the tautology $$A⇒C$$ is, that if $$A$$ is true, then $$C$$ is also true. The truth table proves that the statement $$A ∧ B ⇒ C ∧ B$$ with: $$A⇒C$$ is true

is always true. That means it's correct to write an implication arrow here: $$\frac{20}{x} = \sqrt{(41-x^2 )} ∧ y=\frac{20}{x} ⟹ \frac{400}{x^2} =41-x^2 ∧ y=\frac{20}{x}$$

Again I have analysed the same stuff with the logical OR $$∨$$ : We know that $$A⇒C$$ is true. With this knowledge we analyse the statement $$A ∨ B ⇒ C ∨ B$$ with the help of a truth table:

$$A$$ $$B$$ $$A$$ $$∨$$ $$B$$ $$⟹$$ $$C$$ $$∨$$ $$B$$
0 1 0 1 1 1 ? 1 1
0 0 0 0 0 1 ? ? 0
1 1 1 1 1 1 1 1 1
1 0 1 1 0 1 1 1 0

The truth table proves that the statement $$A ∨ B ⇒ C ∨ B$$ with: $$A⇒C$$ is true

is always true.

Are all these thoughts correct ?

EDIT

@ryang thank you for your answer. Yes I do know about the quantifier ∀x. Maybe I'll write the following to lay some groundwork, so that you also know where I stand with my knowledge: We can think of two predicates $$A(x):(x=2 \Rightarrow x^2=4) \ \ \ \ \mathrm{and} \ \ \ \ B(x):(x^2=4 \Rightarrow x=2 )$$ There are real numbers that turn the predicate $$A(x)$$ into a true statement for eg. $$A(3)$$ is true, $$A(2)$$ is true, $$A(-2)$$ is true. There are real numbers that turn the predicate $$B(x)$$ into a true statement for eg. $$B(3)$$ is true, $$B(2)$$ is true. We can easily see that $$B(-2)$$ is false. Now it turns out that the statement $$\forall x\in \mathbb R: A(x)$$ is a true statement and that the statement $$\forall x\in \mathbb R: B(x)$$ is a false statement (just by looking at the truth value of $$B(-2)$$ ). The reason that we definitely know that $$\forall x\in \mathbb R: A(x)$$ is a true statement is because we know that $$t_1=t_2 \Rightarrow f(t_1)=f(t_2)$$ (with a function $$f$$) is a tautology, because that's just what functions do (input→output). So when we do our usual rearrangings of equations, we actually use functions on both sides of the equation. If the function $$f$$ is injective we can actually put the equivalence arrow $$⇔$$ there. The function used in the predicate $$A(x)$$ looks like this $$f:t\mapsto t^2$$ First I wanted to know, what do you think about the first two truth tables? Are they and the thoughts behind them (how I set them up and how I used them to prove my point) correct?

Regarding your first point, I meant it this way: We know that $$A⇒C$$ is true/a tautology. With this knowledge we analyse the statement $$A ∧ B ⇒ C ∧ B$$ with the help of a truth table.

So I don't think that the truth table you posted is actually correct or maybe I should say, it doesn't represent what I am trying to do. Because the statement $$A⇒C$$ should always have a truth value of 1 since it is a tautology. This should represent the step in the side-calculation where I squared the equation

$$\frac{20}{x} = \sqrt{(41-x^2 )} ⟹ \frac{400}{x^2} =41-x^2.$$

I was trying to prove with a truth table that I can actually write the Implication-Arrow $$⟹$$ here $$\frac{20}{x} = \sqrt{(41-x^2 )} ∧ y=\frac{20}{x} ⟹ \frac{400}{x^2} =41-x^2 ∧ y=\frac{20}{x}.$$

That's why the third and fourth truth table look the way they do. Is it possible to set the truth tables up the way that I did to prove my point?

Regarding your point: "Furthermore, it is not generally valid to determine the truth of a predicate-logic formula by simply dropping its quantifiers and ignoring the internal structure of $$Q(x)$$ by pretending that it is just $$Q$$."

But are we not doing this actually pretty often? For example, let's say I work on the following set of equations

$$y=\frac{1}{6}x^2 \ \ \land \ \ x^2+y^2=16$$ like this $$y=\frac{1}{6}x^2 \ \ \land \ \ x^2+y^2=16 \iff 6y=x^2 \ \ \land \ \ x^2+y^2=16 \iff 6y=x^2 \ \ \land 6y+y^2=16 \\ \iff 6y=x^2 \ \ \land \ \ (y=2 \lor y=-8) \iff 6y=x^2 \ \land \ y=2 \ \ \lor \ \ 6y=x^2 \ \land \ y=-8$$ The reason I knew that this step $$6y=x^2 \ \ \land \ \ (y=2 \lor y=-8) \iff 6y=x^2 \ \land \ y=2 \ \ \lor \ \ 6y=x^2 \ \land \ y=-8$$ is valid, is because earlier I looked at the truth table of $$A \ \land \ (B \lor C) \iff A \ \land \ B \ \ \lor \ \ A \ \land \ C.$$ The truth table shows me that $$A \ \land \ (B \lor C) \iff A \ \land \ B \ \ \lor \ \ A \ \land \ C.$$ is a tautology, so that's why I used this when I worked on my sets of equations.

I hope you see my point.

• For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Commented Aug 13, 2022 at 16:52
• Yes, these implications are all correct, and truth tables are a sensible way of demonstrating them. With your third and fourth tables, given you have $3$ variables, usually a truth table with $2^3 = 8$ rows is used, instead of employing question marks. But, I do follow what you were going for. Commented Aug 13, 2022 at 16:56
• The second half of Using logical implication to analyse mathematics and Addendum 2 of What is a valid step in mathematics? might be of interest. Commented Aug 13, 2022 at 17:44
• @Florian Yes, your thoughts are correct, and yes, I knew what you meant with all of your non-standard truth tables. I wouldn't worry too much about them anyway. Truth tables are primarily a pedagogical tool, designed to anchor abstract concerns about logical arguments in terms of something you can compute. They're bad if you have $4$ or more variables, and basically useless once you introduce quantifiers. You should eventually make more intuitive arguments, and/or use established implications (as theorems). Just use tables to the point where you are convinced by your logical steps. Commented Aug 13, 2022 at 18:01
• For the second table, where you said "I also analysed this with the logical OR: We know that the statement B is false": If the logical OR refers to the only logical OR mentioned before: $\ldots\vee (x-1<0 \wedge -(x-1)>x-2)$, and the statement B refers to its second operand, then this statement B is not always false. Commented Aug 13, 2022 at 18:24

1. The truth table proves that the statement "$$A ∧ B ⇒ C ∧ B$$ with: $$A⇒C$$ is true" is always true.

Does “with” mean ‘if’ or ‘and’? If the former, then you mean $$(A \to C)\to(A ∧ B \to C ∧ B),\tag{*}$$ so why not simply construct this truth table instead? It immediately confirms that $$(*)$$ is a tautology, as required.
$$\begin{array}{ccc|c@{}c@{}ccc@{}ccc@{}c@{}ccc@{}ccc@{}ccc@{}c@{}c@{}c} A&B&C&(&(&A&\rightarrow&C&)&\rightarrow&(&(&A&\land&B&)&\rightarrow&(&C&\land&B&)&)&)\\\hline 1&1&1&&&1&1&1&&\mathbf{1}&&&1&1&1&&1&&1&1&1&&&\\ 1&1&0&&&1&0&0&&\mathbf{1}&&&1&1&1&&0&&0&0&1&&&\\ 1&0&1&&&1&1&1&&\mathbf{1}&&&1&0&0&&1&&1&0&0&&&\\ 1&0&0&&&1&0&0&&\mathbf{1}&&&1&0&0&&1&&0&0&0&&&\\ 0&1&1&&&0&1&1&&\mathbf{1}&&&0&0&1&&1&&1&1&1&&&\\ 0&1&0&&&0&1&0&&\mathbf{1}&&&0&0&1&&1&&0&0&1&&&\\ 0&0&1&&&0&1&1&&\mathbf{1}&&&0&0&0&&1&&1&0&0&&&\\ 0&0&0&&&0&1&0&&\mathbf{1}&&&0&0&0&&1&&0&0&0&&& \end{array}$$

Side note: $$(*)$$ is logically equivalent to $$(A \to C)∧(A ∧ B) \to C ∧ B$$ (because $$P\to(Q\to R)\equiv P\land Q\to R$$); perhaps this is more obviously tautological?

2. That means that $$20/x=\sqrt{41-x^2} ⟹ 400/x^2 =41-x^2$$ implies that $$20/x=\sqrt{41-x^2} ∧ y=20/x ⟹ 400/x^2 =41-x^2 ∧ y=20/x.$$

Let's symbolise this as $$(Ax \to Cx)\to (Ax ∧ Bx\to Cx ∧ Bx).\tag#$$

1. $$(Ax →Cx)$$ is true and $$(*)$$ is a tautology;
2. thus, $$(Ax →Cx)$$ is true and $$(\#)$$ a tautology;
3. thus, $$(Ax ∧ Bx\to Cx ∧ Bx)$$ is true, as required.

we definitely know that is true because we know that is a tautology.

No, the latter is not a tautology, merely universally true for each $$(t_1,t_2).$$

What you mean to say is that the former is true assuming mathematical axioms and the given context, that is, mathematically true.

We know that $$A⇒C$$ is true/a tautology.

$$(A⇒C)$$ is certainly true in your particular context.

But $$(A⇒C)$$ is not a tautology: its truth table has $$4$$ rows, with one $$\text‘0\text’$$ below $$\text‘⇒\text’.$$

With this knowledge, we analyse the statement $$A ∧ B ⇒ C ∧ B$$ with the help of a truth table.

So, you are using $$(A⇒C)$$ as a premise to check whether it is valid to conclude that $$(A∧B⇒C∧B)$$ is true.

In other words, you are checking whether the inference $$(A \to C)\to(A ∧ B \to C ∧ B),\tag{*}$$ is valid (i.e., logically true); in yet other words, you are checking whether $$(*)$$ is a tautology.

So I don't think that the truth table you posted represents what I am trying to do.

My truth table for $$(*)$$ directly demonstrates exactly that $$(A⇒C)$$ tautologically implies $$(A∧B⇒C∧B),$$ as required.

(Observe that it contains $$3$$ variables has $$2^3=8$$ rows, so is a full truth table.)

To wit: let's hunt for any row in which $$(A⇒C)$$ is true and $$(A∧B⇒C∧B)$$ false. (Notice that in this search, we can ignore all rows in which $$(A⇒C)$$ is false, since they won't contain what we're after.) The fact that no such row exists corresponds to the fact that $$(*)$$'s main connective has no $$\text‘0\text’$$ below it.

In a truth table, each row corresponds to each truth assignment (case) of $$(A,B,C).$$ Think of each row as a particular context (‘interpretation’).

In your particular context, $$(A⇒C)$$ is true; however, this is irrelevant: investigating tautological/logical truth means to investigate the truth regardless of context.

Because the statement A⇒C should always have a truth value of 1 since it is a tautology.

Because $$(A⇒C)$$ is not a tautology, a full truth table will contain rows in which $$(A⇒C)$$ has truth value $$0.$$

what do you think about how I set up my truth tables?

Your reasoning process for all those truth tables is valid.

And, using an indirect process and its partial truth table, you have correctly proven that $$(*)$$ is a tautology.

The alternative procedure that I showed is straightforward and mostly mechanical: formalise the given statement, obtain its truth table, then conclude, from the all $$\text‘1\text’$$'s, that it is a tautology.

To be clear: your partial truth table is an excerpt of the truth table that I gave.

• I think that I understood your answer but maybe let me try to say it in my own words: Your truth table shows that each and every time that the sentence $(A⟹C)$ is true, the sentence $(A∧B⟹C∧B)$ is also true. That means, because $$20/x=\sqrt{41-x^2} ⟹ 400/x^2 =41-x^2$$ is true (for all $x∈ \mathbb R$ ), this $$20/x=\sqrt{41-x^2} ∧ y=20/x ⟹ 400/x^2 =41-x^2 ∧ y=20/x$$ is also true (for all $x∈ \mathbb R$). Commented Aug 25, 2022 at 17:46
• No, not with the implicit universal quantification. Invoking universal generalisation at this stage will complexify things (my initial answer had addressed this, it got significantly beyond the level of the OP Question, so I deleted it before posting; I now regret not having just submitted then editing it away, which would've saved a copy for reference). At this point of your analysis, we're merely dealing with an arbitrary $x,$ which is why you can consider just the formulae's truth-functional form and use propositional logic. Commented Aug 25, 2022 at 17:57
• Oh so then I should have written: ... That means, because $$20/x=\sqrt{41-x^2} ⟹ 400/x^2 =41-x^2$$ is true (for an arbitrary $x∈ \mathbb R$), this $$20/x=\sqrt{41-x^2} ∧ y=20/x ⟹ 400/x^2 =41-x^2 ∧ y=20/x$$ is also true (for an arbitrary $x∈ \mathbb R$). Commented Aug 25, 2022 at 18:31
• 1. I guess, but remember, letting $x$ be arbitrary was already declared prior to the current excerpt (that is, it is unseen by us), and there's no need to keep repeating the fact. $\quad$ 2. Instead of belabouring my point from our deleted conversation, Imma just cite Theo's comment. Commented Aug 25, 2022 at 18:41