how to prove that (P⊃Q)≡(¬Q⊃¬P) ( P ⊃ Q ) ≡ ( ¬ Q ⊃ ¬ P ) is disallowed in intuitionistic...
$begingroup$
this is what I've tried;
define kripke model K=({0,1},≤,⊩) where ≤ is the (total) order relation over {0,1} defined by
0≤0 0≤ 11≤1,and ⊩ is a binary relation from {0,1} to the set of propositional variables such that 0⊮A and 1⊩A and 1⊮B and 0⊩B. then
1⊮(A⊃B)≡(¬B⊃¬A) ( B ⊃ A ) ≡ ( ¬ B ⊃ ¬ A )
can somebody tell me if this is correct or if there are errors, what are they and how can they be corrected?
logic intuitionistic-logic
asked Dec 5 '18 at 21:11
katerinekaterine
405
$endgroup$
|
show 2 more comments
$begingroup$
this is what I've tried;
define kripke model K=({0,1},≤,⊩) where ≤ is the (total) order relation over {0,1} defined by
0≤0 0≤ 11≤1,and ⊩ is a binary relation from {0,1} to the set of propositional variables such that 0⊮A and 1⊩A and 1⊮B and 0⊩B. then
1⊮(A⊃B)≡(¬B⊃¬A) ( B ⊃ A ) ≡ ( ¬ B ⊃ ¬ A )
can somebody tell me if this is correct or if there are errors, what are they and how can they be corrected?
logic intuitionistic-logic
asked Dec 5 '18 at 21:11
katerinekaterine
405
$endgroup$
1
$begingroup$
A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:11
$begingroup$
@DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
$endgroup$
– katerine
Dec 5 '18 at 23:33
$begingroup$
I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:37
2
$begingroup$
In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
$endgroup$
– ryan221b
Dec 6 '18 at 0:57
1
$begingroup$
Already asked and answered here.
$endgroup$
– Mauro ALLEGRANZA
Dec 6 '18 at 7:45
|
show 2 more comments
$begingroup$
this is what I've tried;
define kripke model K=({0,1},≤,⊩) where ≤ is the (total) order relation over {0,1} defined by
0≤0 0≤ 11≤1,and ⊩ is a binary relation from {0,1} to the set of propositional variables such that 0⊮A and 1⊩A and 1⊮B and 0⊩B. then
1⊮(A⊃B)≡(¬B⊃¬A) ( B ⊃ A ) ≡ ( ¬ B ⊃ ¬ A )
can somebody tell me if this is correct or if there are errors, what are they and how can they be corrected?
logic intuitionistic-logic
asked Dec 5 '18 at 21:11
katerinekaterine
405
$endgroup$
this is what I've tried;
define kripke model K=({0,1},≤,⊩) where ≤ is the (total) order relation over {0,1} defined by
0≤0 0≤ 11≤1,and ⊩ is a binary relation from {0,1} to the set of propositional variables such that 0⊮A and 1⊩A and 1⊮B and 0⊩B. then
1⊮(A⊃B)≡(¬B⊃¬A) ( B ⊃ A ) ≡ ( ¬ B ⊃ ¬ A )
can somebody tell me if this is correct or if there are errors, what are they and how can they be corrected?
logic intuitionistic-logic
logic intuitionistic-logic
asked Dec 5 '18 at 21:11
katerinekaterine
405
asked Dec 5 '18 at 21:11
katerinekaterine
405
edited Dec 5 '18 at 23:03
katerine
asked Dec 5 '18 at 21:11
katerinekaterine
405
asked Dec 5 '18 at 21:11
katerinekaterine
405
asked Dec 5 '18 at 21:11
katerinekaterine
405
405
1
$begingroup$
A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:11
$begingroup$
@DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
$endgroup$
– katerine
Dec 5 '18 at 23:33
$begingroup$
I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:37
2
$begingroup$
In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
$endgroup$
– ryan221b
Dec 6 '18 at 0:57
1
$begingroup$
Already asked and answered here.
$endgroup$
– Mauro ALLEGRANZA
Dec 6 '18 at 7:45
|
show 2 more comments
1
$begingroup$
A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:11
$begingroup$
@DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
$endgroup$
– katerine
Dec 5 '18 at 23:33
$begingroup$
I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:37
2
$begingroup$
In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
$endgroup$
– ryan221b
Dec 6 '18 at 0:57
1
$begingroup$
Already asked and answered here.
$endgroup$
– Mauro ALLEGRANZA
Dec 6 '18 at 7:45
1
1
$begingroup$
A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:11
$begingroup$
A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:11
$begingroup$
@DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
$endgroup$
– katerine
Dec 5 '18 at 23:33
$begingroup$
@DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
$endgroup$
– katerine
Dec 5 '18 at 23:33
$begingroup$
I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:37
$begingroup$
I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:37
2
2
$begingroup$
In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
$endgroup$
– ryan221b
Dec 6 '18 at 0:57
$begingroup$
In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
$endgroup$
– ryan221b
Dec 6 '18 at 0:57
1
1
$begingroup$
Already asked and answered here.
$endgroup$
– Mauro ALLEGRANZA
Dec 6 '18 at 7:45
$begingroup$
Already asked and answered here.
$endgroup$
– Mauro ALLEGRANZA
Dec 6 '18 at 7:45
|
show 2 more comments
1 Answer
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$begingroup$
I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.
Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.
I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:
- Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.
- On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.
Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.
You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!
answered Dec 6 '18 at 2:01
ryan221bryan221b
9510
$endgroup$
add a comment |
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$begingroup$
I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.
Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.
I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:
- Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.
- On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.
Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.
You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!
answered Dec 6 '18 at 2:01
ryan221bryan221b
9510
$endgroup$
add a comment |
$begingroup$
I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.
Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.
I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:
- Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.
- On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.
Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.
You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!
answered Dec 6 '18 at 2:01
ryan221bryan221b
9510
$endgroup$
add a comment |
$begingroup$
I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.
Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.
I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:
- Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.
- On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.
Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.
You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!
answered Dec 6 '18 at 2:01
ryan221bryan221b
9510
$endgroup$
I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.
Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.
I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:
- Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.
- On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.
Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.
You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!
answered Dec 6 '18 at 2:01
ryan221bryan221b
9510
answered Dec 6 '18 at 2:01
ryan221bryan221b
9510
answered Dec 6 '18 at 2:01
ryan221bryan221b
9510
answered Dec 6 '18 at 2:01
ryan221bryan221b
9510
9510
add a comment |
add a comment |
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1
$begingroup$
A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:11
$begingroup$
@DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
$endgroup$
– katerine
Dec 5 '18 at 23:33
$begingroup$
I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:37
2
$begingroup$
In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
$endgroup$
– ryan221b
Dec 6 '18 at 0:57
1
$begingroup$
Already asked and answered here.
$endgroup$
– Mauro ALLEGRANZA
Dec 6 '18 at 7:45