Distributive property of tensor ($otimes$) over par (⅋) in linear logic
$begingroup$
In the setting of linear logic, does $otimes$ distribute over $⅋$?
That is, is it possible to show that
$$
A otimes (B ⅋ C) stackrel?equiv
(A otimes B) ⅋ (A otimes C)
$$
holds? If not, what is a counterexample?
The intuition in terms of resources is that if I have $A$ and a multiplicative disjunction of $B$ and $C$ it is the same as saying that I have a multiplicative disjunction of ($A otimes B$) and ($Aotimes B$).
Other distributive properties of linear logic. For reference, it is certainly true that $otimes$ distributes over $oplus$ and that $⅋$ distributes over $&$:
$$ A otimes (B oplus C) equiv (A otimes B)oplus (Aotimes C)$$
$$ A ⅋ (B & C) equiv (A ⅋ B) & (A ⅋ C)$$
tensor-products linear-logic
asked Jan 13 at 18:06
Andrea CensiAndrea Censi
113
$endgroup$
add a comment |
$begingroup$
In the setting of linear logic, does $otimes$ distribute over $⅋$?
That is, is it possible to show that
$$
A otimes (B ⅋ C) stackrel?equiv
(A otimes B) ⅋ (A otimes C)
$$
holds? If not, what is a counterexample?
The intuition in terms of resources is that if I have $A$ and a multiplicative disjunction of $B$ and $C$ it is the same as saying that I have a multiplicative disjunction of ($A otimes B$) and ($Aotimes B$).
Other distributive properties of linear logic. For reference, it is certainly true that $otimes$ distributes over $oplus$ and that $⅋$ distributes over $&$:
$$ A otimes (B oplus C) equiv (A otimes B)oplus (Aotimes C)$$
$$ A ⅋ (B & C) equiv (A ⅋ B) & (A ⅋ C)$$
tensor-products linear-logic
asked Jan 13 at 18:06
Andrea CensiAndrea Censi
113
$endgroup$
3
$begingroup$
Sorry, what is ⅋?
$endgroup$
– Berci
Jan 13 at 18:11
$begingroup$
Something not that intuitive, as this previous discussion shows. See here for the formal definition using sequents.
$endgroup$
– Andrea Censi
Jan 13 at 18:14
add a comment |
$begingroup$
In the setting of linear logic, does $otimes$ distribute over $⅋$?
That is, is it possible to show that
$$
A otimes (B ⅋ C) stackrel?equiv
(A otimes B) ⅋ (A otimes C)
$$
holds? If not, what is a counterexample?
The intuition in terms of resources is that if I have $A$ and a multiplicative disjunction of $B$ and $C$ it is the same as saying that I have a multiplicative disjunction of ($A otimes B$) and ($Aotimes B$).
Other distributive properties of linear logic. For reference, it is certainly true that $otimes$ distributes over $oplus$ and that $⅋$ distributes over $&$:
$$ A otimes (B oplus C) equiv (A otimes B)oplus (Aotimes C)$$
$$ A ⅋ (B & C) equiv (A ⅋ B) & (A ⅋ C)$$
tensor-products linear-logic
asked Jan 13 at 18:06
Andrea CensiAndrea Censi
113
$endgroup$
In the setting of linear logic, does $otimes$ distribute over $⅋$?
That is, is it possible to show that
$$
A otimes (B ⅋ C) stackrel?equiv
(A otimes B) ⅋ (A otimes C)
$$
holds? If not, what is a counterexample?
The intuition in terms of resources is that if I have $A$ and a multiplicative disjunction of $B$ and $C$ it is the same as saying that I have a multiplicative disjunction of ($A otimes B$) and ($Aotimes B$).
Other distributive properties of linear logic. For reference, it is certainly true that $otimes$ distributes over $oplus$ and that $⅋$ distributes over $&$:
$$ A otimes (B oplus C) equiv (A otimes B)oplus (Aotimes C)$$
$$ A ⅋ (B & C) equiv (A ⅋ B) & (A ⅋ C)$$
tensor-products linear-logic
tensor-products linear-logic
asked Jan 13 at 18:06
Andrea CensiAndrea Censi
113
asked Jan 13 at 18:06
Andrea CensiAndrea Censi
113
edited Jan 13 at 18:15
Andrea Censi
asked Jan 13 at 18:06
Andrea CensiAndrea Censi
113
asked Jan 13 at 18:06
Andrea CensiAndrea Censi
113
asked Jan 13 at 18:06
Andrea CensiAndrea Censi
113
113
3
$begingroup$
Sorry, what is ⅋?
$endgroup$
– Berci
Jan 13 at 18:11
$begingroup$
Something not that intuitive, as this previous discussion shows. See here for the formal definition using sequents.
$endgroup$
– Andrea Censi
Jan 13 at 18:14
add a comment |
3
$begingroup$
Sorry, what is ⅋?
$endgroup$
– Berci
Jan 13 at 18:11
$begingroup$
Something not that intuitive, as this previous discussion shows. See here for the formal definition using sequents.
$endgroup$
– Andrea Censi
Jan 13 at 18:14
3
3
$begingroup$
Sorry, what is ⅋?
$endgroup$
– Berci
Jan 13 at 18:11
$begingroup$
Sorry, what is ⅋?
$endgroup$
– Berci
Jan 13 at 18:11
$begingroup$
Something not that intuitive, as this previous discussion shows. See here for the formal definition using sequents.
$endgroup$
– Andrea Censi
Jan 13 at 18:14
$begingroup$
Something not that intuitive, as this previous discussion shows. See here for the formal definition using sequents.
$endgroup$
– Andrea Censi
Jan 13 at 18:14
add a comment |
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$begingroup$
Sorry, what is ⅋?
$endgroup$
– Berci
Jan 13 at 18:11
$begingroup$
Something not that intuitive, as this previous discussion shows. See here for the formal definition using sequents.
$endgroup$
– Andrea Censi
Jan 13 at 18:14