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.
Best Answer
Following your clarification EDIT, I have revised my Answer correspondingly, and addressed your new questions as an Addendum below.
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?
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!
ADDENDUM
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.
$(A⇒C)$ 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 $\text‘⇒\text’.$
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.
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 $(A⇒C)$ is not a tautology, a full truth table will contain rows in which $(A⇒C)$ has truth value $0.$
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.