Reductio ad absurdum, but just once












4












$begingroup$


Can we prove that $exists x lnot F$ follows from $lnot forall x F$ by natural deduction using the intuitionistic rules, and RAA (reductio ad absurdum) just once?



It is usually done by using RAA twice on the assumptions $lnot F$ and $lnot exists x lnot F$. I know that $forall x F$ doesn't follow from $lnot exists x lnot F$ in intuitionistic logic, so we cannot derive $exists x lnot F$ from $forall x F, lnot forall x F$. Moreover, if we introduced $exists x lnot F$ by RAA, we would have to derive $bot$ from $lnot exists x lnot F, lnot forall x F$; if we introduced $exists x lnot F$ by $exists$-intro, we would have to derive $lnot F$ from $lnot forall x F$ using RAA just once, which seems not quite likely... In both cases I don't know how to proceed.










share|cite|improve this question









$endgroup$












  • $begingroup$
    It seems to me that it shouldn't be possible if you're just using RAA on $exists x. neg F$. It's a bit fuzzy for me, but if I recall correctly, the existence property for intuitionistic FOL (since we've already used our one alotted instance of RAA, we're working in intutionistic logic from this point onward) should imply that $neg forall x. F, neg exists x. neg F notvdash exists x. neg F$. Of course, that doesn't rule out a more creative use of RAA, but I don't see how it can be done.
    $endgroup$
    – user181407
    Feb 11 '17 at 21:45
















4












$begingroup$


Can we prove that $exists x lnot F$ follows from $lnot forall x F$ by natural deduction using the intuitionistic rules, and RAA (reductio ad absurdum) just once?



It is usually done by using RAA twice on the assumptions $lnot F$ and $lnot exists x lnot F$. I know that $forall x F$ doesn't follow from $lnot exists x lnot F$ in intuitionistic logic, so we cannot derive $exists x lnot F$ from $forall x F, lnot forall x F$. Moreover, if we introduced $exists x lnot F$ by RAA, we would have to derive $bot$ from $lnot exists x lnot F, lnot forall x F$; if we introduced $exists x lnot F$ by $exists$-intro, we would have to derive $lnot F$ from $lnot forall x F$ using RAA just once, which seems not quite likely... In both cases I don't know how to proceed.










share|cite|improve this question









$endgroup$












  • $begingroup$
    It seems to me that it shouldn't be possible if you're just using RAA on $exists x. neg F$. It's a bit fuzzy for me, but if I recall correctly, the existence property for intuitionistic FOL (since we've already used our one alotted instance of RAA, we're working in intutionistic logic from this point onward) should imply that $neg forall x. F, neg exists x. neg F notvdash exists x. neg F$. Of course, that doesn't rule out a more creative use of RAA, but I don't see how it can be done.
    $endgroup$
    – user181407
    Feb 11 '17 at 21:45














4












4








4


1



$begingroup$


Can we prove that $exists x lnot F$ follows from $lnot forall x F$ by natural deduction using the intuitionistic rules, and RAA (reductio ad absurdum) just once?



It is usually done by using RAA twice on the assumptions $lnot F$ and $lnot exists x lnot F$. I know that $forall x F$ doesn't follow from $lnot exists x lnot F$ in intuitionistic logic, so we cannot derive $exists x lnot F$ from $forall x F, lnot forall x F$. Moreover, if we introduced $exists x lnot F$ by RAA, we would have to derive $bot$ from $lnot exists x lnot F, lnot forall x F$; if we introduced $exists x lnot F$ by $exists$-intro, we would have to derive $lnot F$ from $lnot forall x F$ using RAA just once, which seems not quite likely... In both cases I don't know how to proceed.










share|cite|improve this question









$endgroup$




Can we prove that $exists x lnot F$ follows from $lnot forall x F$ by natural deduction using the intuitionistic rules, and RAA (reductio ad absurdum) just once?



It is usually done by using RAA twice on the assumptions $lnot F$ and $lnot exists x lnot F$. I know that $forall x F$ doesn't follow from $lnot exists x lnot F$ in intuitionistic logic, so we cannot derive $exists x lnot F$ from $forall x F, lnot forall x F$. Moreover, if we introduced $exists x lnot F$ by RAA, we would have to derive $bot$ from $lnot exists x lnot F, lnot forall x F$; if we introduced $exists x lnot F$ by $exists$-intro, we would have to derive $lnot F$ from $lnot forall x F$ using RAA just once, which seems not quite likely... In both cases I don't know how to proceed.







logic natural-deduction intuitionistic-logic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 7 '17 at 10:17









kylekyle

916




916












  • $begingroup$
    It seems to me that it shouldn't be possible if you're just using RAA on $exists x. neg F$. It's a bit fuzzy for me, but if I recall correctly, the existence property for intuitionistic FOL (since we've already used our one alotted instance of RAA, we're working in intutionistic logic from this point onward) should imply that $neg forall x. F, neg exists x. neg F notvdash exists x. neg F$. Of course, that doesn't rule out a more creative use of RAA, but I don't see how it can be done.
    $endgroup$
    – user181407
    Feb 11 '17 at 21:45


















  • $begingroup$
    It seems to me that it shouldn't be possible if you're just using RAA on $exists x. neg F$. It's a bit fuzzy for me, but if I recall correctly, the existence property for intuitionistic FOL (since we've already used our one alotted instance of RAA, we're working in intutionistic logic from this point onward) should imply that $neg forall x. F, neg exists x. neg F notvdash exists x. neg F$. Of course, that doesn't rule out a more creative use of RAA, but I don't see how it can be done.
    $endgroup$
    – user181407
    Feb 11 '17 at 21:45
















$begingroup$
It seems to me that it shouldn't be possible if you're just using RAA on $exists x. neg F$. It's a bit fuzzy for me, but if I recall correctly, the existence property for intuitionistic FOL (since we've already used our one alotted instance of RAA, we're working in intutionistic logic from this point onward) should imply that $neg forall x. F, neg exists x. neg F notvdash exists x. neg F$. Of course, that doesn't rule out a more creative use of RAA, but I don't see how it can be done.
$endgroup$
– user181407
Feb 11 '17 at 21:45




$begingroup$
It seems to me that it shouldn't be possible if you're just using RAA on $exists x. neg F$. It's a bit fuzzy for me, but if I recall correctly, the existence property for intuitionistic FOL (since we've already used our one alotted instance of RAA, we're working in intutionistic logic from this point onward) should imply that $neg forall x. F, neg exists x. neg F notvdash exists x. neg F$. Of course, that doesn't rule out a more creative use of RAA, but I don't see how it can be done.
$endgroup$
– user181407
Feb 11 '17 at 21:45










1 Answer
1






active

oldest

votes


















3












$begingroup$

You cannot derive $lnotforall x F(x) rightarrow exists xlnot F(x)$ in an intuitionistic system without RAA. This is because there is a Soundness Theorem for intuitionistic logic with respect to Kripke Semantics (https://en.wikipedia.org/wiki/Kripke_semantics#Semantics_of_intuitionistic_logic; I spell them out a bit below): if we could derive that implication using the rules (without RAA), then the formula would be valid in all Kripke Models. However, we can construct a counter-example in which $lnotforall x F(x) rightarrow exists xlnot F(x)$ does not hold, so it cannot be derived.



Consider the following Kripke Structure $mathcal{K} = (K, leq, D, Vdash)$:



enter image description here



Since $0leq1$, $1notVdash F(b)$ and $bin D(1)$, we have $0notVdashforall x F(x)$, and also $1notVdashforall x F(x)$, so $0Vdash lnotforall x F(x)$. But at 0, there is no $din D(0)$ satisfying $lnot F(d)$, so $0notVdashexists x lnot F(x)$. Hence, $0notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$.





In case you're unfamiliar with Kripke Structures. $(K, leq)$ is a partially ordered set where you can consider the elements of $K$ to be "states of knowledge": at each stage $kin K$ we "know" some theorems but perhaps not others, which may be proved or refuted at a later stage $k' geq k$. Each state has a domain of objects, $D(k)$, and $Vdash$ tells us which sentences hold at each node.



Two observations:




  1. Objects do not "disappear" once constructed: $k leq k' implies D(k) subset D(k')$.

  2. Proved theorems remain proved: $k Vdashvarphi implies k' Vdashvarphi$ for all $k' geq k$.


(You can easily see that the example above satisfies these conditions.)



We can now stipulate the conditions under which $k Vdashvarphi$:





  • $kVdash F(d) iff$ $F(d)$ holds at $k$ and $din D(k)$;


  • $kVdash lnotvarphi iff$ for all $k'geq k$, $k'notVdashvarphi$;


  • $kVdash varphilandpsi iff kVdashvarphi text{ and } kVdashpsi$;


  • $kVdash varphilorpsi iff kVdashvarphi text{ or } kVdashpsi$;


  • $k Vdash varphirightarrowpsi iff $ for all $k' geq k$, $k'Vdashvarphi$ implies $k'Vdashpsi$;


  • $kVdashforall x F(x) iff $ for all $k' geq k$ and all $din D(k')$, $k'Vdash F(d)$;


  • $kVdashexists x F(x) iff $ there is a $din D(k)$ such that $kVdash F(d)$;


We say that a formula $varphi$ is Kripke valid if, for all Kripke Structures $mathcal{K} = (K, D, Vdash)$ and all $kin K$, $kVdashvarphi$, and we write "$Vdashvarphi$". Writing "$vdash_ivarphi$" for "$varphi$ is intuitionistically provable", we have




Soundness Theorem If $vdash_ivarphi$, then $Vdashvarphi$.




Hence, since we showed that $notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$ (by providing a Kripke Structure as a counter-example), it follows from the Soundness Theorem that $notvdash_ilnotforall x F(x) rightarrow exists x lnot F(x)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I interpreted the question regarding "only one RAA" as meaning: could you prove something along the lines of $lnot forall x, F(x) rightarrow lnotlnot exists x, lnot F(x)$. (For which my counterexample in a Heyting algebra model - which I'm much more familiar with personally - would be: Heyting algebra is open subsets of $mathbb{R}$, domain of discourse is $mathbb{R}$, $F(x) = mathbb{R} setminus { x }$.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 0:32










  • $begingroup$
    In the above structure we also have $0Vdashlnotlnotexists x lnot F(x)$, so it serves as a counter-example to both implications.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 2:04










  • $begingroup$
    If $0 Vdash lnotlnot exists x lnot F(x)$ then how is it a counterexample to $lnot forall x F(x) rightarrow lnotlnot exists x lnot F(x)$?
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 4:26










  • $begingroup$
    I meant $0notVdashlnotlnotexists x not F(x)$, but I don't think that actually holds in my example now, I'll need a different counterexample.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 11:11






  • 1




    $begingroup$
    How about: $mathbb{P} = (mathbb{N}, le)$; $D(k) = mathbb{N}$ for all $k$; and $k Vdash F(n)$ if and only if $k ge n$. If I'm not mistaken, in this example, $0 Vdash lnot forall x F(x)$ but $0 notVdash lnotlnot exists x lnot F(x)$ (and in fact $0 Vdash lnot exists x lnot F(x)$). But I'm not that confident in my computations here...
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 18:12











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3












$begingroup$

You cannot derive $lnotforall x F(x) rightarrow exists xlnot F(x)$ in an intuitionistic system without RAA. This is because there is a Soundness Theorem for intuitionistic logic with respect to Kripke Semantics (https://en.wikipedia.org/wiki/Kripke_semantics#Semantics_of_intuitionistic_logic; I spell them out a bit below): if we could derive that implication using the rules (without RAA), then the formula would be valid in all Kripke Models. However, we can construct a counter-example in which $lnotforall x F(x) rightarrow exists xlnot F(x)$ does not hold, so it cannot be derived.



Consider the following Kripke Structure $mathcal{K} = (K, leq, D, Vdash)$:



enter image description here



Since $0leq1$, $1notVdash F(b)$ and $bin D(1)$, we have $0notVdashforall x F(x)$, and also $1notVdashforall x F(x)$, so $0Vdash lnotforall x F(x)$. But at 0, there is no $din D(0)$ satisfying $lnot F(d)$, so $0notVdashexists x lnot F(x)$. Hence, $0notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$.





In case you're unfamiliar with Kripke Structures. $(K, leq)$ is a partially ordered set where you can consider the elements of $K$ to be "states of knowledge": at each stage $kin K$ we "know" some theorems but perhaps not others, which may be proved or refuted at a later stage $k' geq k$. Each state has a domain of objects, $D(k)$, and $Vdash$ tells us which sentences hold at each node.



Two observations:




  1. Objects do not "disappear" once constructed: $k leq k' implies D(k) subset D(k')$.

  2. Proved theorems remain proved: $k Vdashvarphi implies k' Vdashvarphi$ for all $k' geq k$.


(You can easily see that the example above satisfies these conditions.)



We can now stipulate the conditions under which $k Vdashvarphi$:





  • $kVdash F(d) iff$ $F(d)$ holds at $k$ and $din D(k)$;


  • $kVdash lnotvarphi iff$ for all $k'geq k$, $k'notVdashvarphi$;


  • $kVdash varphilandpsi iff kVdashvarphi text{ and } kVdashpsi$;


  • $kVdash varphilorpsi iff kVdashvarphi text{ or } kVdashpsi$;


  • $k Vdash varphirightarrowpsi iff $ for all $k' geq k$, $k'Vdashvarphi$ implies $k'Vdashpsi$;


  • $kVdashforall x F(x) iff $ for all $k' geq k$ and all $din D(k')$, $k'Vdash F(d)$;


  • $kVdashexists x F(x) iff $ there is a $din D(k)$ such that $kVdash F(d)$;


We say that a formula $varphi$ is Kripke valid if, for all Kripke Structures $mathcal{K} = (K, D, Vdash)$ and all $kin K$, $kVdashvarphi$, and we write "$Vdashvarphi$". Writing "$vdash_ivarphi$" for "$varphi$ is intuitionistically provable", we have




Soundness Theorem If $vdash_ivarphi$, then $Vdashvarphi$.




Hence, since we showed that $notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$ (by providing a Kripke Structure as a counter-example), it follows from the Soundness Theorem that $notvdash_ilnotforall x F(x) rightarrow exists x lnot F(x)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I interpreted the question regarding "only one RAA" as meaning: could you prove something along the lines of $lnot forall x, F(x) rightarrow lnotlnot exists x, lnot F(x)$. (For which my counterexample in a Heyting algebra model - which I'm much more familiar with personally - would be: Heyting algebra is open subsets of $mathbb{R}$, domain of discourse is $mathbb{R}$, $F(x) = mathbb{R} setminus { x }$.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 0:32










  • $begingroup$
    In the above structure we also have $0Vdashlnotlnotexists x lnot F(x)$, so it serves as a counter-example to both implications.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 2:04










  • $begingroup$
    If $0 Vdash lnotlnot exists x lnot F(x)$ then how is it a counterexample to $lnot forall x F(x) rightarrow lnotlnot exists x lnot F(x)$?
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 4:26










  • $begingroup$
    I meant $0notVdashlnotlnotexists x not F(x)$, but I don't think that actually holds in my example now, I'll need a different counterexample.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 11:11






  • 1




    $begingroup$
    How about: $mathbb{P} = (mathbb{N}, le)$; $D(k) = mathbb{N}$ for all $k$; and $k Vdash F(n)$ if and only if $k ge n$. If I'm not mistaken, in this example, $0 Vdash lnot forall x F(x)$ but $0 notVdash lnotlnot exists x lnot F(x)$ (and in fact $0 Vdash lnot exists x lnot F(x)$). But I'm not that confident in my computations here...
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 18:12
















3












$begingroup$

You cannot derive $lnotforall x F(x) rightarrow exists xlnot F(x)$ in an intuitionistic system without RAA. This is because there is a Soundness Theorem for intuitionistic logic with respect to Kripke Semantics (https://en.wikipedia.org/wiki/Kripke_semantics#Semantics_of_intuitionistic_logic; I spell them out a bit below): if we could derive that implication using the rules (without RAA), then the formula would be valid in all Kripke Models. However, we can construct a counter-example in which $lnotforall x F(x) rightarrow exists xlnot F(x)$ does not hold, so it cannot be derived.



Consider the following Kripke Structure $mathcal{K} = (K, leq, D, Vdash)$:



enter image description here



Since $0leq1$, $1notVdash F(b)$ and $bin D(1)$, we have $0notVdashforall x F(x)$, and also $1notVdashforall x F(x)$, so $0Vdash lnotforall x F(x)$. But at 0, there is no $din D(0)$ satisfying $lnot F(d)$, so $0notVdashexists x lnot F(x)$. Hence, $0notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$.





In case you're unfamiliar with Kripke Structures. $(K, leq)$ is a partially ordered set where you can consider the elements of $K$ to be "states of knowledge": at each stage $kin K$ we "know" some theorems but perhaps not others, which may be proved or refuted at a later stage $k' geq k$. Each state has a domain of objects, $D(k)$, and $Vdash$ tells us which sentences hold at each node.



Two observations:




  1. Objects do not "disappear" once constructed: $k leq k' implies D(k) subset D(k')$.

  2. Proved theorems remain proved: $k Vdashvarphi implies k' Vdashvarphi$ for all $k' geq k$.


(You can easily see that the example above satisfies these conditions.)



We can now stipulate the conditions under which $k Vdashvarphi$:





  • $kVdash F(d) iff$ $F(d)$ holds at $k$ and $din D(k)$;


  • $kVdash lnotvarphi iff$ for all $k'geq k$, $k'notVdashvarphi$;


  • $kVdash varphilandpsi iff kVdashvarphi text{ and } kVdashpsi$;


  • $kVdash varphilorpsi iff kVdashvarphi text{ or } kVdashpsi$;


  • $k Vdash varphirightarrowpsi iff $ for all $k' geq k$, $k'Vdashvarphi$ implies $k'Vdashpsi$;


  • $kVdashforall x F(x) iff $ for all $k' geq k$ and all $din D(k')$, $k'Vdash F(d)$;


  • $kVdashexists x F(x) iff $ there is a $din D(k)$ such that $kVdash F(d)$;


We say that a formula $varphi$ is Kripke valid if, for all Kripke Structures $mathcal{K} = (K, D, Vdash)$ and all $kin K$, $kVdashvarphi$, and we write "$Vdashvarphi$". Writing "$vdash_ivarphi$" for "$varphi$ is intuitionistically provable", we have




Soundness Theorem If $vdash_ivarphi$, then $Vdashvarphi$.




Hence, since we showed that $notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$ (by providing a Kripke Structure as a counter-example), it follows from the Soundness Theorem that $notvdash_ilnotforall x F(x) rightarrow exists x lnot F(x)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I interpreted the question regarding "only one RAA" as meaning: could you prove something along the lines of $lnot forall x, F(x) rightarrow lnotlnot exists x, lnot F(x)$. (For which my counterexample in a Heyting algebra model - which I'm much more familiar with personally - would be: Heyting algebra is open subsets of $mathbb{R}$, domain of discourse is $mathbb{R}$, $F(x) = mathbb{R} setminus { x }$.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 0:32










  • $begingroup$
    In the above structure we also have $0Vdashlnotlnotexists x lnot F(x)$, so it serves as a counter-example to both implications.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 2:04










  • $begingroup$
    If $0 Vdash lnotlnot exists x lnot F(x)$ then how is it a counterexample to $lnot forall x F(x) rightarrow lnotlnot exists x lnot F(x)$?
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 4:26










  • $begingroup$
    I meant $0notVdashlnotlnotexists x not F(x)$, but I don't think that actually holds in my example now, I'll need a different counterexample.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 11:11






  • 1




    $begingroup$
    How about: $mathbb{P} = (mathbb{N}, le)$; $D(k) = mathbb{N}$ for all $k$; and $k Vdash F(n)$ if and only if $k ge n$. If I'm not mistaken, in this example, $0 Vdash lnot forall x F(x)$ but $0 notVdash lnotlnot exists x lnot F(x)$ (and in fact $0 Vdash lnot exists x lnot F(x)$). But I'm not that confident in my computations here...
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 18:12














3












3








3





$begingroup$

You cannot derive $lnotforall x F(x) rightarrow exists xlnot F(x)$ in an intuitionistic system without RAA. This is because there is a Soundness Theorem for intuitionistic logic with respect to Kripke Semantics (https://en.wikipedia.org/wiki/Kripke_semantics#Semantics_of_intuitionistic_logic; I spell them out a bit below): if we could derive that implication using the rules (without RAA), then the formula would be valid in all Kripke Models. However, we can construct a counter-example in which $lnotforall x F(x) rightarrow exists xlnot F(x)$ does not hold, so it cannot be derived.



Consider the following Kripke Structure $mathcal{K} = (K, leq, D, Vdash)$:



enter image description here



Since $0leq1$, $1notVdash F(b)$ and $bin D(1)$, we have $0notVdashforall x F(x)$, and also $1notVdashforall x F(x)$, so $0Vdash lnotforall x F(x)$. But at 0, there is no $din D(0)$ satisfying $lnot F(d)$, so $0notVdashexists x lnot F(x)$. Hence, $0notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$.





In case you're unfamiliar with Kripke Structures. $(K, leq)$ is a partially ordered set where you can consider the elements of $K$ to be "states of knowledge": at each stage $kin K$ we "know" some theorems but perhaps not others, which may be proved or refuted at a later stage $k' geq k$. Each state has a domain of objects, $D(k)$, and $Vdash$ tells us which sentences hold at each node.



Two observations:




  1. Objects do not "disappear" once constructed: $k leq k' implies D(k) subset D(k')$.

  2. Proved theorems remain proved: $k Vdashvarphi implies k' Vdashvarphi$ for all $k' geq k$.


(You can easily see that the example above satisfies these conditions.)



We can now stipulate the conditions under which $k Vdashvarphi$:





  • $kVdash F(d) iff$ $F(d)$ holds at $k$ and $din D(k)$;


  • $kVdash lnotvarphi iff$ for all $k'geq k$, $k'notVdashvarphi$;


  • $kVdash varphilandpsi iff kVdashvarphi text{ and } kVdashpsi$;


  • $kVdash varphilorpsi iff kVdashvarphi text{ or } kVdashpsi$;


  • $k Vdash varphirightarrowpsi iff $ for all $k' geq k$, $k'Vdashvarphi$ implies $k'Vdashpsi$;


  • $kVdashforall x F(x) iff $ for all $k' geq k$ and all $din D(k')$, $k'Vdash F(d)$;


  • $kVdashexists x F(x) iff $ there is a $din D(k)$ such that $kVdash F(d)$;


We say that a formula $varphi$ is Kripke valid if, for all Kripke Structures $mathcal{K} = (K, D, Vdash)$ and all $kin K$, $kVdashvarphi$, and we write "$Vdashvarphi$". Writing "$vdash_ivarphi$" for "$varphi$ is intuitionistically provable", we have




Soundness Theorem If $vdash_ivarphi$, then $Vdashvarphi$.




Hence, since we showed that $notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$ (by providing a Kripke Structure as a counter-example), it follows from the Soundness Theorem that $notvdash_ilnotforall x F(x) rightarrow exists x lnot F(x)$.






share|cite|improve this answer











$endgroup$



You cannot derive $lnotforall x F(x) rightarrow exists xlnot F(x)$ in an intuitionistic system without RAA. This is because there is a Soundness Theorem for intuitionistic logic with respect to Kripke Semantics (https://en.wikipedia.org/wiki/Kripke_semantics#Semantics_of_intuitionistic_logic; I spell them out a bit below): if we could derive that implication using the rules (without RAA), then the formula would be valid in all Kripke Models. However, we can construct a counter-example in which $lnotforall x F(x) rightarrow exists xlnot F(x)$ does not hold, so it cannot be derived.



Consider the following Kripke Structure $mathcal{K} = (K, leq, D, Vdash)$:



enter image description here



Since $0leq1$, $1notVdash F(b)$ and $bin D(1)$, we have $0notVdashforall x F(x)$, and also $1notVdashforall x F(x)$, so $0Vdash lnotforall x F(x)$. But at 0, there is no $din D(0)$ satisfying $lnot F(d)$, so $0notVdashexists x lnot F(x)$. Hence, $0notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$.





In case you're unfamiliar with Kripke Structures. $(K, leq)$ is a partially ordered set where you can consider the elements of $K$ to be "states of knowledge": at each stage $kin K$ we "know" some theorems but perhaps not others, which may be proved or refuted at a later stage $k' geq k$. Each state has a domain of objects, $D(k)$, and $Vdash$ tells us which sentences hold at each node.



Two observations:




  1. Objects do not "disappear" once constructed: $k leq k' implies D(k) subset D(k')$.

  2. Proved theorems remain proved: $k Vdashvarphi implies k' Vdashvarphi$ for all $k' geq k$.


(You can easily see that the example above satisfies these conditions.)



We can now stipulate the conditions under which $k Vdashvarphi$:





  • $kVdash F(d) iff$ $F(d)$ holds at $k$ and $din D(k)$;


  • $kVdash lnotvarphi iff$ for all $k'geq k$, $k'notVdashvarphi$;


  • $kVdash varphilandpsi iff kVdashvarphi text{ and } kVdashpsi$;


  • $kVdash varphilorpsi iff kVdashvarphi text{ or } kVdashpsi$;


  • $k Vdash varphirightarrowpsi iff $ for all $k' geq k$, $k'Vdashvarphi$ implies $k'Vdashpsi$;


  • $kVdashforall x F(x) iff $ for all $k' geq k$ and all $din D(k')$, $k'Vdash F(d)$;


  • $kVdashexists x F(x) iff $ there is a $din D(k)$ such that $kVdash F(d)$;


We say that a formula $varphi$ is Kripke valid if, for all Kripke Structures $mathcal{K} = (K, D, Vdash)$ and all $kin K$, $kVdashvarphi$, and we write "$Vdashvarphi$". Writing "$vdash_ivarphi$" for "$varphi$ is intuitionistically provable", we have




Soundness Theorem If $vdash_ivarphi$, then $Vdashvarphi$.




Hence, since we showed that $notVdashlnotforall x F(x) rightarrow exists xlnot F(x)$ (by providing a Kripke Structure as a counter-example), it follows from the Soundness Theorem that $notvdash_ilnotforall x F(x) rightarrow exists x lnot F(x)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 20:05

























answered Dec 6 '18 at 0:07









ryan221bryan221b

10510




10510












  • $begingroup$
    I interpreted the question regarding "only one RAA" as meaning: could you prove something along the lines of $lnot forall x, F(x) rightarrow lnotlnot exists x, lnot F(x)$. (For which my counterexample in a Heyting algebra model - which I'm much more familiar with personally - would be: Heyting algebra is open subsets of $mathbb{R}$, domain of discourse is $mathbb{R}$, $F(x) = mathbb{R} setminus { x }$.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 0:32










  • $begingroup$
    In the above structure we also have $0Vdashlnotlnotexists x lnot F(x)$, so it serves as a counter-example to both implications.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 2:04










  • $begingroup$
    If $0 Vdash lnotlnot exists x lnot F(x)$ then how is it a counterexample to $lnot forall x F(x) rightarrow lnotlnot exists x lnot F(x)$?
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 4:26










  • $begingroup$
    I meant $0notVdashlnotlnotexists x not F(x)$, but I don't think that actually holds in my example now, I'll need a different counterexample.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 11:11






  • 1




    $begingroup$
    How about: $mathbb{P} = (mathbb{N}, le)$; $D(k) = mathbb{N}$ for all $k$; and $k Vdash F(n)$ if and only if $k ge n$. If I'm not mistaken, in this example, $0 Vdash lnot forall x F(x)$ but $0 notVdash lnotlnot exists x lnot F(x)$ (and in fact $0 Vdash lnot exists x lnot F(x)$). But I'm not that confident in my computations here...
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 18:12


















  • $begingroup$
    I interpreted the question regarding "only one RAA" as meaning: could you prove something along the lines of $lnot forall x, F(x) rightarrow lnotlnot exists x, lnot F(x)$. (For which my counterexample in a Heyting algebra model - which I'm much more familiar with personally - would be: Heyting algebra is open subsets of $mathbb{R}$, domain of discourse is $mathbb{R}$, $F(x) = mathbb{R} setminus { x }$.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 0:32










  • $begingroup$
    In the above structure we also have $0Vdashlnotlnotexists x lnot F(x)$, so it serves as a counter-example to both implications.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 2:04










  • $begingroup$
    If $0 Vdash lnotlnot exists x lnot F(x)$ then how is it a counterexample to $lnot forall x F(x) rightarrow lnotlnot exists x lnot F(x)$?
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 4:26










  • $begingroup$
    I meant $0notVdashlnotlnotexists x not F(x)$, but I don't think that actually holds in my example now, I'll need a different counterexample.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 11:11






  • 1




    $begingroup$
    How about: $mathbb{P} = (mathbb{N}, le)$; $D(k) = mathbb{N}$ for all $k$; and $k Vdash F(n)$ if and only if $k ge n$. If I'm not mistaken, in this example, $0 Vdash lnot forall x F(x)$ but $0 notVdash lnotlnot exists x lnot F(x)$ (and in fact $0 Vdash lnot exists x lnot F(x)$). But I'm not that confident in my computations here...
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 18:12
















$begingroup$
I interpreted the question regarding "only one RAA" as meaning: could you prove something along the lines of $lnot forall x, F(x) rightarrow lnotlnot exists x, lnot F(x)$. (For which my counterexample in a Heyting algebra model - which I'm much more familiar with personally - would be: Heyting algebra is open subsets of $mathbb{R}$, domain of discourse is $mathbb{R}$, $F(x) = mathbb{R} setminus { x }$.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 0:32




$begingroup$
I interpreted the question regarding "only one RAA" as meaning: could you prove something along the lines of $lnot forall x, F(x) rightarrow lnotlnot exists x, lnot F(x)$. (For which my counterexample in a Heyting algebra model - which I'm much more familiar with personally - would be: Heyting algebra is open subsets of $mathbb{R}$, domain of discourse is $mathbb{R}$, $F(x) = mathbb{R} setminus { x }$.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 0:32












$begingroup$
In the above structure we also have $0Vdashlnotlnotexists x lnot F(x)$, so it serves as a counter-example to both implications.
$endgroup$
– ryan221b
Dec 6 '18 at 2:04




$begingroup$
In the above structure we also have $0Vdashlnotlnotexists x lnot F(x)$, so it serves as a counter-example to both implications.
$endgroup$
– ryan221b
Dec 6 '18 at 2:04












$begingroup$
If $0 Vdash lnotlnot exists x lnot F(x)$ then how is it a counterexample to $lnot forall x F(x) rightarrow lnotlnot exists x lnot F(x)$?
$endgroup$
– Daniel Schepler
Dec 6 '18 at 4:26




$begingroup$
If $0 Vdash lnotlnot exists x lnot F(x)$ then how is it a counterexample to $lnot forall x F(x) rightarrow lnotlnot exists x lnot F(x)$?
$endgroup$
– Daniel Schepler
Dec 6 '18 at 4:26












$begingroup$
I meant $0notVdashlnotlnotexists x not F(x)$, but I don't think that actually holds in my example now, I'll need a different counterexample.
$endgroup$
– ryan221b
Dec 6 '18 at 11:11




$begingroup$
I meant $0notVdashlnotlnotexists x not F(x)$, but I don't think that actually holds in my example now, I'll need a different counterexample.
$endgroup$
– ryan221b
Dec 6 '18 at 11:11




1




1




$begingroup$
How about: $mathbb{P} = (mathbb{N}, le)$; $D(k) = mathbb{N}$ for all $k$; and $k Vdash F(n)$ if and only if $k ge n$. If I'm not mistaken, in this example, $0 Vdash lnot forall x F(x)$ but $0 notVdash lnotlnot exists x lnot F(x)$ (and in fact $0 Vdash lnot exists x lnot F(x)$). But I'm not that confident in my computations here...
$endgroup$
– Daniel Schepler
Dec 6 '18 at 18:12




$begingroup$
How about: $mathbb{P} = (mathbb{N}, le)$; $D(k) = mathbb{N}$ for all $k$; and $k Vdash F(n)$ if and only if $k ge n$. If I'm not mistaken, in this example, $0 Vdash lnot forall x F(x)$ but $0 notVdash lnotlnot exists x lnot F(x)$ (and in fact $0 Vdash lnot exists x lnot F(x)$). But I'm not that confident in my computations here...
$endgroup$
– Daniel Schepler
Dec 6 '18 at 18:12


















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