I don't get how "universal generalization" works? is my understanding of UI, EI, EG. correct?

Question

So, I think I understand how UI, EI, EG work.

• Universal instantiation:

Let's suppose that there's a set $A=\{a_1,a_2,a_3, ...,a_n\}$ such that $\forall_x[x∈A⇒φ(x)]$

This means that

$a_1$ fulfills $φ(x)$

$a_2$ fulfills $φ(x)$

$...$

$a_i$ fulfills $φ(x)|1≤i≤n$

$...$

$a_n$ fulfills $φ(x)$

So I can take any member $a_w|1≤w≤n$, and $a_w∈A⇒φ(a_w)$; which is what one wants, making $\forall_x[x∈A⇒φ(x)]$ a simple propositional logic statement.

• Existential instantiation:

Let's suppose that there's a set $A=\{a_1,a_2,a_3, ...,a_n\}$ such that $\exists_x[x∈A∧φ(x)]$

So I know there's at least $1$ member of $A$ that fulfills $φ(x)$, I'm going to call it $a_k$, so $a_k∈A∧φ(a_k)$. This is desirable, since we want to turn $\exists_x[x∈A∧φ(x)]$ into a propositional logic statement.

• Existential generalization:

Let's suppose that there's a set $A=\{a_1,a_2,a_3, ...,a_n\}$ and I don't know anything about it; so exploring it, I realize that $φ(a_z)$ is true, I mean, $a_z$ fulfills a property $φ(x)$, or more formally, $a_z∈A∧φ(a_z)$. I know there's at least one member of $A$ that fulfills property $φ(x)$, there could be more, or maybe this is the only one, but I can conclude $\exists_x[x∈A∧φ(x)]$, which is desirable, since we want to turn a propositional logic statment, into a first-order logic quantified statement.

• Universal generalization.

This is the one I don't get.

According to what I have understood, it goes like this, and I know it's really bad explained.

let's suppose I have "a thing" which is susceptible to be the member of a set, and we call it $w$, so somehow I find out $φ(w)$ is true, and later on I find out $w∈A$, so $w∈A∧φ(w)$; therefore, I can conclude $\forall_x[x∈A⇒φ(x)]$.

I don't think this is actually how UG works, and if it is, I don't how to apply it to real life, because according to this, I can just take a ball out of a bag, and if it's blue, then conclude all of them are blue, which is not necessarily true.

I would be really grateful if you explained to me how this rule works, I have tried to understand it for days.

Thanks in advance.

Answer 1

Existential instantiation :

Let's suppose that there's a set $A = \{ a_1,a_2,a_3,...,a_n \}$ such that $∃x[x∈A ∧ φ(x)]$ holds.

So I know there's at least one member of $A$ that fulfills $φ(x)$, I'm going to call it $a_k$.

Not exactly : we do not know which of $A$'s fulfills $φ(x)$. Thus, we add a "temporary name" $c$ (a new name, not previously used in the proof : it can be a new variable) for the unknown element of $A$ whose existence we know for sure and we are licensed to assert that $φ(c)$ holds.

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Universal generalization :

Let assume that we have proved for a "generic" element $x \in A$ that $φ(x)$ holds.

Then, we can freely conclude that $φ(x)$ holds of every element of $A$, i.e. that $∀x[x∈A \to φ(x)]$ holds.

The trick is in the "generic" requirement: do not assume any specific properties about $x$ but only the fact that it is an element of $A$.

If we do this, we can conclude that whatever we are able to prove about $x$ will hold for every object that has the same "properties" of $x$: i.e. to belong to $A$.

Answer 2

In mathematical reasoning we often want to argue that a property holds "for all $x$" (in some understood domain). The way we often go about this is we consider an arbitrary $x,$ that has no properties other than those it must have by virtue of being an element of the understood domain. Then we show that the property holds for $x.$ Since we did not assume anything additional about $x,$ we reason that means this property must hold for all $x.$ Universal generalization is a formalization of this line of reasoning. It allows us to pass from having proved an open formula that includes a variable (the variable is our "arbitrary element") to a closed for all statement.

Answer 3

Let's suppose that there's a set $A=\{a_1,~a_2,~a_3,~\dots,~a_n\}$ and I don't know anything about it

For existential generalization, the second part isn't necessary at all. You can know anything about $A$ and you can have lots of assumptions about the $a_z$. Since all you are concluding is "there is something", knowing things about the something is fine.

let's suppose I have "a thing" which is susceptible to be the member of a set, and we call it $w$

You are confusing yourself because you are trying to describe UGEN in terms of sets. That's like trying to learn to walk while juggling.

UGEN is simply this: if you can prove $P(x)$, and you have not assumed anything about $x$, then you can conclude $\forall x.P(x)$.

Adding sets to the above statement means you are trying to do 2 things at once. For example, suppose $A$ is all multiples of 6 : $\{0, 6, 12, \dots\}$. Suppose $Q(x)$ is "$x$ is even".

When you consider an $x$ such that $x \in A$, then $x$ is not arbitrary. When you consider $x$ such that $x$ is even, it is still not arbitrary. But when you generalize the $\to$ into $x \in A \to Q(x)$, then $x$ is arbitrary: you can prove the previous statement without assuming anything about $x$.

I don't how to apply it to real life

Your friend asked you "If someone accidentally broke your phone, would you be mad at them?" You said "Yes". Your friend will then know that, because "someone" could be "anyone", he can apply UGEN and conclude "for all persons $x$, if $x$ accidentally broke you phone, you will be mad at $x$". He will then apply UI and conclude "if I accidentally broke your phone, you will be mad at me".

On the other hand,

Your friend asked you "If your girlfriend accidentally broke your windshield, would you be mad at her?" You say "no". Your friend cannot apply UGEN: girlfriend is not an arbitrary person. You might be mad at him if he breaks your windshield.