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Following Prof.Mummert suggestion about the correct application of Intro- and Elimination- rules for quantifiers [pag.98-99]:

look at examples 5 and 6 on page 149 of Kleene 1952.

Example 5. Given $A(x) \vdash B$ and $A(x) \vdash \lnot B$ with x held constant, then $\vdash \lnot A(x)$ and $\vdash \forall x \lnot A(x)$. Supply the formal details.

1) $A(x) \vdash B$

2) $A(x) \vdash \lnot B$

applying ($\lnot-I$) [i.e. if $\Gamma, A \vdash B$ and $\Gamma,A \vdash \lnot B$ then $\Gamma \vdash \lnot A$ with $\Gamma = \emptyset$ ] to 1 and 2 :

3) $\vdash \lnot A(x)$

then apply ($\forall$-I) to 3 :

4) $\vdash \forall \lnot xA(x)$.

Another proof :

1) $A(x) \vdash B$

2) $\vdash A(x) \rightarrow B$ --- from 1,($\rightarrow$-I)

3) $A(x) \vdash \lnot B$

4) $\vdash A(x) \rightarrow \lnot B$ --- from 3,($\rightarrow$-I)

5) $\vdash (A \rightarrow B) \rightarrow (A \rightarrow \lnot B) \rightarrow \lnot A$ --- Axiom 7

7) $\vdash \lnot A(x)$ --- from 2,5 and 4,6 with two applications of ($\rightarrow$-E)

8) $\vdash \forall \lnot xA(x)$ --- from 7,($\forall$-I).

Example 6. Given $A(x) \vdash^x B$ and $A(x) \vdash^x \lnot B$ with x not necessarly held constant, then $\vdash \lnot \forall x A(x)$. Supply the formal details.

Mauro ALLEGRANZA
  • 99,247

1 Answers1

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Example 6 is stated in term of $\vdash^x$.

Following the "intuitive meaning" of this relation, we would have :

from 1) $\forall xA(x) \vdash B$ --- with ($\rightarrow$-I)

and 2) $\forall xA(x) \vdash \lnot B$ --- with ($\rightarrow$-I)

and using Axiom 7 : $\vdash (A \rightarrow B) \rightarrow (A \rightarrow \lnot B) \rightarrow \lnot A$

3) $\vdash \lnot \forall xA(x)$ --- with two applications of ($\rightarrow$-E).

But in Kleene's system we have no formal rule that licenses "from $A(x) \vdash^x B$ to $\forall xA(x) \vdash B$".

The correct solution is :

1) $A(x) \vdash^x B$

2) $\forall xA(x)$ --- Assumption

3) $A(x)$ --- from 2, ($\forall$-E)

now, having derived $A(x)$, using the properties of $\vdash$ [rif pag.89] :

4) $\forall xA(x) \vdash^x B$

5) $\vdash \forall xA(x) \rightarrow B$ --- from 4, ($\rightarrow$)-I

6) $A(x) \vdash^x \lnot B$

in the same way :

10) $\vdash \forall xA(x) \rightarrow \lnot B$

now, from 5,10 and Axiom 7 :

12) $\vdash \lnot \forall xA(x)$ --- two applications of ($\rightarrow$)-E.

Mauro ALLEGRANZA
  • 99,247