Which first order logic logic equivalences are not valid in intuitionistic logic?
0
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I know that --A == A and -(A / B) == -A or -B don't hold in intuitionistic logic. Which are the relatives first order logic equivalence that don't hold in intuitionistic logic?
first-order-logic
asked Jan 4 at 11:11
MaicakeMaicake
716
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add a comment |
0
$begingroup$
I know that --A == A and -(A / B) == -A or -B don't hold in intuitionistic logic. Which are the relatives first order logic equivalence that don't hold in intuitionistic logic?
first-order-logic
asked Jan 4 at 11:11
MaicakeMaicake
716
$endgroup$
$begingroup$
See Intuitionistic logic.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:20
2
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For example, it is not true that $lnot forall x lnot Px$ is equivalent to $exists x Px$.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:23
2
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There are lots of examples. For examples, some kind of de Morgan's law $lnot (Aland B)to lnot Alor lnot B$ is not valid. $Ato B$ no longer implies $lnot Alor B$. You may find some examples by considering a statement under BHK interpretation.
$endgroup$
– Hanul Jeon
Jan 4 at 11:30
add a comment |
0
0
0
$begingroup$
I know that --A == A and -(A / B) == -A or -B don't hold in intuitionistic logic. Which are the relatives first order logic equivalence that don't hold in intuitionistic logic?
first-order-logic
asked Jan 4 at 11:11
MaicakeMaicake
716
$endgroup$
I know that --A == A and -(A / B) == -A or -B don't hold in intuitionistic logic. Which are the relatives first order logic equivalence that don't hold in intuitionistic logic?
first-order-logic
first-order-logic
asked Jan 4 at 11:11
MaicakeMaicake
716
asked Jan 4 at 11:11
MaicakeMaicake
716
asked Jan 4 at 11:11
MaicakeMaicake
716
asked Jan 4 at 11:11
MaicakeMaicake
716
asked Jan 4 at 11:11
MaicakeMaicake
716
716
$begingroup$
See Intuitionistic logic.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:20
2
$begingroup$
For example, it is not true that $lnot forall x lnot Px$ is equivalent to $exists x Px$.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:23
2
$begingroup$
There are lots of examples. For examples, some kind of de Morgan's law $lnot (Aland B)to lnot Alor lnot B$ is not valid. $Ato B$ no longer implies $lnot Alor B$. You may find some examples by considering a statement under BHK interpretation.
$endgroup$
– Hanul Jeon
Jan 4 at 11:30
add a comment |
$begingroup$
See Intuitionistic logic.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:20
2
$begingroup$
For example, it is not true that $lnot forall x lnot Px$ is equivalent to $exists x Px$.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:23
2
$begingroup$
There are lots of examples. For examples, some kind of de Morgan's law $lnot (Aland B)to lnot Alor lnot B$ is not valid. $Ato B$ no longer implies $lnot Alor B$. You may find some examples by considering a statement under BHK interpretation.
$endgroup$
– Hanul Jeon
Jan 4 at 11:30
$begingroup$
See Intuitionistic logic.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:20
$begingroup$
See Intuitionistic logic.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:20
2
2
$begingroup$
For example, it is not true that $lnot forall x lnot Px$ is equivalent to $exists x Px$.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:23
$begingroup$
For example, it is not true that $lnot forall x lnot Px$ is equivalent to $exists x Px$.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:23
2
2
$begingroup$
There are lots of examples. For examples, some kind of de Morgan's law $lnot (Aland B)to lnot Alor lnot B$ is not valid. $Ato B$ no longer implies $lnot Alor B$. You may find some examples by considering a statement under BHK interpretation.
$endgroup$
– Hanul Jeon
Jan 4 at 11:30
$begingroup$
There are lots of examples. For examples, some kind of de Morgan's law $lnot (Aland B)to lnot Alor lnot B$ is not valid. $Ato B$ no longer implies $lnot Alor B$. You may find some examples by considering a statement under BHK interpretation.
$endgroup$
– Hanul Jeon
Jan 4 at 11:30
add a comment |
0
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$begingroup$
See Intuitionistic logic.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:20
2
$begingroup$
For example, it is not true that $lnot forall x lnot Px$ is equivalent to $exists x Px$.
$endgroup$
– Mauro ALLEGRANZA
Jan 4 at 11:23
2
$begingroup$
There are lots of examples. For examples, some kind of de Morgan's law $lnot (Aland B)to lnot Alor lnot B$ is not valid. $Ato B$ no longer implies $lnot Alor B$. You may find some examples by considering a statement under BHK interpretation.
$endgroup$
– Hanul Jeon
Jan 4 at 11:30