Equations and truth tables

Question

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 ?

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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.

Answer 1

Following your clarification EDIT, I have revised my Answer correspondingly, and addressed your new questions as an Addendum below.

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 $$(Ax \to Cx)\to (Ax ∧ Bx\to Cx ∧ Bx).\tag#$$

   Here's your argument, made explicit. Yes, it is sound!

   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.

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ADDENDUM

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.