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Math Help - Want to negate the following statements

  1. #31
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    Re: Want to negate the following statements

    The universal introduction rule

    \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, a is a constant, so you can't apply this rule to derive \forall x\, P(x).

    On the other hand, \neg\exists x\,\neg P(x) and \neg P(a) are inconsistent and therefore do imply any formula, including \forall x\,P(x): one can conclude any formula directly from \bot, which is obtained from \neg\exists x\neg P(x) and \exists x\neg P(x). However, that arbitrary conclusion would still have an open assumption \neg P(a). In contrast, the proof above takes care to close the temporary assumption \neg P(x).
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  2. #32
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    Re: Want to negate the following statements

    Quote Originally Posted by emakarov View Post
    In your derivation, a is a constant, so you can't apply this rule to derive \forall x\, P(x).

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

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

    I.COPI ,page 73 ,in his Fourth edition ,Symbolic Logic
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  3. #33
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    Re: Want to negate the following statements

    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.
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  4. #34
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    Re: Want to negate the following statements

    Quote Originally Posted by emakarov View Post
    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
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  5. #35
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    Re: Want to negate the following statements

    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).

    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.
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  6. #36
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    Re: Want to negate the following statements

    Quote Originally Posted by emakarov View Post
    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 :

    \neg Pa\Longrightarrow\forall x Px
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  7. #37
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    Re: Want to negate the following statements

    Quote Originally Posted by alexandros View Post
    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 :

    \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.
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