Re: Want to negate the following statements

The universal introduction rule

$\displaystyle \frac{P(x)}{\forall x\,P(x)}$

requires that x in the premise is a free variable. (In addition, it must not be free in any open assumption of P(x).) In your derivation, $\displaystyle a$ is a constant, so you can't apply this rule to derive $\displaystyle \forall x\, P(x)$.

On the other hand, $\displaystyle \neg\exists x\,\neg P(x)$ and $\displaystyle \neg P(a)$ are inconsistent and therefore do imply any formula, including $\displaystyle \forall x\,P(x)$: one can conclude any formula directly from $\displaystyle \bot$, which is obtained from $\displaystyle \neg\exists x\neg P(x)$ and $\displaystyle \exists x\neg P(x)$. However, that arbitrary conclusion would still have an open assumption $\displaystyle \neg P(a)$. In contrast, the proof above takes care to close the temporary assumption $\displaystyle \neg P(x)$.

Re: Want to negate the following statements

Quote:

Originally Posted by

**emakarov** In your derivation, $\displaystyle a$ is a constant, so you can't apply this rule to derive $\displaystyle \forall x\, P(x)$.

.

In my derivation a is any arbritary selected individual (Antony in my case).

HENCE I can apply the generalization rule $\displaystyle \forall x\, P(x)$.

I.COPI ,page 73 ,in his Fourth edition ,Symbolic Logic

Re: Want to negate the following statements

Quote:

n my derivation a is any arbritary selected individual (Antony in my case).

Unfortunately, I don't have a copy of Copi's book. There are two options: either *a* is a object variable, or *a* is a constant. This is how first-order terms are built by definition.

If *a* is a variable, then your derivation in post #30 is correct. It derives ∀x P(x) from an open assumption ~∃x ~P(x), so this is not surprising.

Re: Want to negate the following statements

Quote:

Originally Posted by

**emakarov** Unfortunately, I don't have a copy of Copi's book. There are two options: either

*a* is a object variable, or

*a* is a constant. This is how first-order terms are built by definition.

If

*a* is a variable, then your derivation in

post #30 is correct. It derives ∀x P(x) from an open assumption ~∃x ~P(x), so this is not surprising.

In my post #30 i derived ∀x P(x) from ~Pa ,hence the sentence:

if Antony is not phonny ,then everybody is a phony ,which is not true

Re: Want to negate the following statements

Quote:

In my post #30 i derived ∀x P(x) from ~Pa

We need to be clear about the assumptions of the derivation of ∀x P(x). It essentially uses the assumption ~∃x ~Px to derive a contradiction. On the other hand, it does not have ~Pa as an open assumption: it was closed when ~~P(a) was derived from a contradiction. The negation introduction rule, which is used at that point, takes a derivation of a contradiction from some open assumption A and derives ~A; in the process it *closes* A. Therefore, your derivation of ∀x P(x) in post #30 has a single open assumption: ~∃x ~Px, i.e., essentially, it is a derivation of (~∃x ~Px) -> ∀x P(x).

Quote:

if Antony is not phonny ,then everybody is a phony ,which is not true

It is somewhat misleading to interpret ~Pa as "Antony is not phony" when *a* is a free variable. When you say that *a* stands for a particular person Antony, you treat *a* as a constant that has a fixed meaning in your interpretation. Formulas with free variables in general cannot be interpreted as a propositions; their truth value depends on the value of the free variables.

Re: Want to negate the following statements

Quote:

Originally Posted by

**emakarov** We need to be clear about the assumptions of the derivation of ∀x P(x). It essentially uses the assumption ~∃x ~Px to derive a contradiction. On the other hand, it does not have ~Pa as an open assumption: it was closed when ~~P(a) was derived from a contradiction. The negation introduction rule, which is used at that point, takes a derivation of a contradiction from some open assumption A and derives ~A; in the process it *closes* A. Therefore, your derivation of ∀x P(x) in post #30 has a single open assumption: ~∃x ~Px, i.e., essentially, it is a derivation of (~∃x ~Px) -> ∀x P(x).

In page 56 line 13 from the top, of Angelo's Margaris FIRST ORDER MATHEMATICAL LOGIC we read and i quote:

"When an assumption is discharged (or closed to use your terminology)by the deduction theorem it does not disappear; it is transferred across |= (|= is a Logical symbol symbolizing conclusion) to become the antecedent of a conditional "

Hence in our case ~Pa ,when discharged ,it does not disappear , it is transferred further down the proof to become the antecedent of a conditional ,which is :

$\displaystyle \neg Pa\Longrightarrow\forall x Px$

Re: Want to negate the following statements

Quote:

Originally Posted by

**alexandros** In page 56 line 13 from the top, of Angelo's Margaris FIRST ORDER MATHEMATICAL LOGIC we read and i quote:

"When an assumption is discharged (or closed to use your terminology)by the deduction theorem it does not disappear; it is transferred across |= (|= is a Logical symbol symbolizing conclusion) to become the antecedent of a conditional "

Hence in our case ~Pa ,when discharged ,it does not disappear , it is transferred further down the proof to become the antecedent of a conditional ,which is :

$\displaystyle \neg Pa\Longrightarrow\forall x Px$

The thread has gone quite off-topic to what the OP posted. I can see no end in sight. Thread closed.