Logic: Universal/Existential Generalization After Assumption

Question

I am having a problem working out the quantifiers in the following context. Any help would be greatly appreciated. I have set this problem up kind of like Fitch notation.

Assume $(\forall x, x\in A) \rightarrow x\in B$.

Case 1: Assume $c\in A$.

Thus, by line 1 (universal instantiation), $c\in A \rightarrow c\in B$.

Thus, by line 2 and 3 (modus ponens) $c\in B$.

Am I okay to use universal generalization (from line 4) here or do I have to use existential generalization? In other words, would saying $\forall x, x\in B$ by universal generalization be problematic in this context?

Case 2: Assume $c\notin A$.

Again, by line 1 (universal instantiation), $c\in A\rightarrow c\in B$.

As $c\notin A$ is assumed true, then $c\in A$ must be false. Thus, we know $c\in B$. Then, I am stuck at the same part in bold shown above.

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I am trying to prove the proposition: $\forall a \in G$, $a\in center(G)$ $\leftrightarrow$ $\forall x\in G$, $a\in C(x)$.

In this context, $G$ is a group, $center(G)=\lbrace x| \forall y\in G, x=y^{-1}xy \wedge x\in G\rbrace$, and $C(x)=\lbrace y|y=x^{-1}yx \wedge y\in G\rbrace$ for some $x\in G$.

In context to the logic, I am just going to post how I was thinking of doing the problem:

$\rightarrow$ Assume $\forall a, a \in G\ \to a\in center(G)$. Thus, by universal instantiation, $a\in G \rightarrow a\in center(G)$ for an arbitrary $a$. Assume $a\notin G$. Thus, $a\in G$ would be false; in other words, we would have a contradiction. Thus, we know $a\in G$ by negation introduction. So, $a\in center(G)$ by modus ponens as we stated previously that $a\in G \rightarrow a\in center(G)$. Now, as $a\in center(G)$, we know $\forall x\in G, a=x^{-1}ax \wedge a\in G$. In other words, $\forall x\in G$, $a\in C(x)$. By universal instantiation, we know $\forall a\in G$, $\forall x\in G$, $a\in C(x)$. Here the quantifiers don't match up.

$\leftarrow$ Assume $\forall a\in G$, $\forall x\in G$, $a\in C(x)$. In other words, assume $\forall x\in G$, $\forall a\in G$, $ax=xa$. Again, in other words, $\forall a\in G$, $a\in Center(G)$. This is what we wanted to show.

Answer 1

You really could use some parentheses to gain some clarity. Apparently, what you need to show is:

$\forall a \in G \big( a\in center(G)$ $\leftrightarrow$ $\forall x\in G$, $a\in C(x)\big)$

So, the whole statement is a universal, but it is a universal of a biconditional, and as far as the biconditional itself is concerned, you have a universal on one side, but not on the other side. So, things are a little more complicated than what you are trying to do. Try following this basic proof format 'plan':

Answer 2

I am trying to prove the proposition: $\forall a \in G$, $a\in center(G)$ $\leftrightarrow$ $\forall x\in G$, $a\in C(x)$.

In this case, I prefer this notation:

$\forall a: [a \in G \implies[a\in center(G) \iff\forall b:[b\in G \implies a\in C(b)]]$

The required structure of your proof should be more evident with this notation.

Hint: The first 3 lines of your proof would be something like this:

1. Suppose $x\in G\space\space$ (Premise)

2. Suppose $x\in center(G)\space\space$ (Premise)

3. Suppose $y\in G\space\space$ (Premise)

Next, prove that $x\in C(y)$.

Then generalize to obtain $\forall b:[b\in G \implies x\in C(b)]$ and you will be halfway there.