Can I do instantiation like this?

Question

Suppose, if I have been given this: $\forall x \in A(P(x))$ and $\exists y \in A(Q(y))$.

Now from $\forall x \in A(P(x))$ using universal instantiation, I get $P(c)$ where $c$ is an arbitrary element in $A$. Now my doubt is, if using $\exists y \in A(Q(y))$, can I conclude $Q(c)$ ?. My reasoning says yes, because $c$ is an arbitrary element in $A$ and $y \in A$. Is this type of instantiation from universal to existential has a special name ?

Answer 1

No, you have to go the other way around: from $\exists y\in A(Q(y))$ we instantiate to get a $c$ with $Q(c)$; we now conclude that $P(c)$ holds, because $\forall x\in A(P(c))$.

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One way to picture this is as a game. You're trying to show that a statement - say, $\exists x\in A(P(x) \wedge Q(x))$ - is true, and your opponent (me) is trying to thwart you. Your example goes as follows:

• You: "I know that $\forall x\in A(P(x))$. Give me some $x\in A$ such that $P(x)$."

• Me: "Sure, have this thing $c$. I promise you $P(c)$ holds."

• You: "I know that there is some $x$ with $Q(x)$. I claim $Q(c)$ holds, in particular, so I've found an $x$ with $P(x)\wedge Q(x)$."

• Me: "Haha, I chose a $c$ that doesn't satisfy $Q(c)$! You have made an incorrect deduction, so you lose."

The right way to approach this is the opposite:

• You: "I know $\exists x\in A(Q(x))$. Give me some $x\in A$ with $Q(x)$."

• Me: "Sure, have this thing $c$. I promise you $Q(c)$ holds."

• You: "Well, I know that $\forall x\in A(P(x))$ - so in particular, $P(c)$. So I have $P(c)\wedge Q(c)$."

• Me: "Curses!"

The general name for this sort of thing is game semantics. I find it a very helpful framework for understanding the (in)validity of logical arguments (and many other things besides).

Answer 2

Now from $∀x∈A(P(x))$ using universal instantiation, I get $P(c)$ where $c$ is an arbitrary element in A . Now my doubt is, if using $∃y∈A(Q(y))$ , can I conclude $Q(c)$ ?.

No.

My reasoning says yes, because $c$ is an arbitrary element in $A$ and $y∈A$ . Is this type of instantiation from universal to existential has a special name ?

Your reasoning is almost correct.   You can pick a value $c$ where $Q(c)$ holds, but this not an arbitrary value from a universal instantiation of the first statement.

It's a witness from an existential instantiation of the second statement.

You can then select this witness from a universal instantiation, but, as it's not arbitrary you cannot subsequently use it in universal generalisation.   A witness can only be used in a subsequent existential generalisation.

Thus we do so:

$\begin{align} \exists y\in A\;Q(y) \;&\vdash\; Q(c)&\text{EI}, c\in A \\[2ex] \forall x\in A\;P(x) \;&\vdash\; P(c)&\text{UI}(\mid_c^x) \\[1ex] P(c), Q(c) \;&\vdash\; P(c)\wedge Q(c)&\wedge\text{I} \\[1ex] P(c)\wedge Q(c) \;&\vdash\; \exists x\in A\;(P(x)\wedge Q(x))&\text{EG} \\[2ex] \therefore \forall x\in A\;P(x),\, \exists y\in A\;Q(y)\;&\vdash\; \exists x\in A\;(P(x)\wedge Q(x)) & \Box \end{align}$