How to prove that in intuitionistic logic the contrapositive law is disallowed?
here are my efforts:
I need to show that $(P supset Q) equiv (neg Q supset neg P)$
for this I need to construct a kripke model. I construct the following model:
$K = ({0,1}, leq , models)$
then I define the followings for $A$ and $B$
$0 nvDash A$ $1 vDash A$
$ 1 nvDash B$ $0 vDash B $
then I say that
$ 1 nvDash (neg B to neg A) to Ato B$
Can somebody say what is wrong with my proof?
logic
edited Nov 16 at 20:21
Mauro ALLEGRANZA
64.1k448111
asked Nov 16 at 20:18
jadenperesl
11
add a comment |
here are my efforts:
I need to show that $(P supset Q) equiv (neg Q supset neg P)$
for this I need to construct a kripke model. I construct the following model:
$K = ({0,1}, leq , models)$
then I define the followings for $A$ and $B$
$0 nvDash A$ $1 vDash A$
$ 1 nvDash B$ $0 vDash B $
then I say that
$ 1 nvDash (neg B to neg A) to Ato B$
Can somebody say what is wrong with my proof?
logic
edited Nov 16 at 20:21
Mauro ALLEGRANZA
64.1k448111
asked Nov 16 at 20:18
jadenperesl
11
1
Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
– spaceisdarkgreen
Nov 16 at 21:41
add a comment |
here are my efforts:
I need to show that $(P supset Q) equiv (neg Q supset neg P)$
for this I need to construct a kripke model. I construct the following model:
$K = ({0,1}, leq , models)$
then I define the followings for $A$ and $B$
$0 nvDash A$ $1 vDash A$
$ 1 nvDash B$ $0 vDash B $
then I say that
$ 1 nvDash (neg B to neg A) to Ato B$
Can somebody say what is wrong with my proof?
logic
edited Nov 16 at 20:21
Mauro ALLEGRANZA
64.1k448111
asked Nov 16 at 20:18
jadenperesl
11
here are my efforts:
I need to show that $(P supset Q) equiv (neg Q supset neg P)$
for this I need to construct a kripke model. I construct the following model:
$K = ({0,1}, leq , models)$
then I define the followings for $A$ and $B$
$0 nvDash A$ $1 vDash A$
$ 1 nvDash B$ $0 vDash B $
then I say that
$ 1 nvDash (neg B to neg A) to Ato B$
Can somebody say what is wrong with my proof?
logic
logic
edited Nov 16 at 20:21
Mauro ALLEGRANZA
64.1k448111
asked Nov 16 at 20:18
jadenperesl
11
edited Nov 16 at 20:21
Mauro ALLEGRANZA
64.1k448111
asked Nov 16 at 20:18
jadenperesl
11
edited Nov 16 at 20:21
Mauro ALLEGRANZA
64.1k448111
edited Nov 16 at 20:21
Mauro ALLEGRANZA
64.1k448111
edited Nov 16 at 20:21
Mauro ALLEGRANZA
64.1k448111
64.1k448111
asked Nov 16 at 20:18
jadenperesl
11
asked Nov 16 at 20:18
jadenperesl
11
asked Nov 16 at 20:18
jadenperesl
11
11
1
Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
– spaceisdarkgreen
Nov 16 at 21:41
add a comment |
1
Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
– spaceisdarkgreen
Nov 16 at 21:41
1
1
Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
– spaceisdarkgreen
Nov 16 at 21:41
Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
– spaceisdarkgreen
Nov 16 at 21:41
add a comment |
1 Answer
1
active
oldest
votes
Long comment
You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.
We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.
In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.
answered Nov 29 at 12:25
Mauro ALLEGRANZA
64.1k448111
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
Long comment
You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.
We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.
In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.
answered Nov 29 at 12:25
Mauro ALLEGRANZA
64.1k448111
add a comment |
Long comment
You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.
We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.
In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.
answered Nov 29 at 12:25
Mauro ALLEGRANZA
64.1k448111
add a comment |
Long comment
You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.
We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.
In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.
answered Nov 29 at 12:25
Mauro ALLEGRANZA
64.1k448111
Long comment
You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.
We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.
In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.
answered Nov 29 at 12:25
Mauro ALLEGRANZA
64.1k448111
answered Nov 29 at 12:25
Mauro ALLEGRANZA
64.1k448111
answered Nov 29 at 12:25
Mauro ALLEGRANZA
64.1k448111
answered Nov 29 at 12:25
Mauro ALLEGRANZA
64.1k448111
64.1k448111
add a comment |
add a comment |
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1
Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
– spaceisdarkgreen
Nov 16 at 21:41