I find it stated, in all the resources I have searched, that the following De Morgan laws$$(A\otimes B)^{\perp}\equiv A^{\perp}\wp B^{\perp}\quad\quad\quad (A\text{&}B)^{\perp}\equiv A^\perp \oplus B^\perp$$$$(A\wp B)^{\perp}\equiv A^{\perp}\otimes B^{\perp}\quad\quad\quad (A\oplus B)^{\perp}\equiv A^\perp \text{&} B^\perp$$where, as I universally find in literature, $\text{&}$ is the additive conjunction, $\otimes$ is the multiplicative conjunction, $\oplus$ is the additive disjunction, $\wp$ is the multiplicative disjunction and $^\perp$ is the linear negation, hold in linear logic.

***EDIT*** **Mar 14'15**: I had not found a proof of such equivalences and I were not able to produce one for myself to understand them, since, although I knew the [sequent rules](http://plato.stanford.edu/entries/logic-linear/#SeqCal) used in linear logic, I had never found a precise description of how derivations work in such a system, until the [answer](https://math.stackexchange.com/questions/1179954/de-morgan-laws-of-linear-logic/1180691#1180691) kindly given below explained to me how to derive $(A\otimes B)^\bot\vdash A^\bot\wp B^\bot$, $(A\wp B)^\bot\vdash A^\bot\otimes B^\bot$, $(A\oplus B)^\bot\vdash A^\bot\& B^\bot$ and $(A\& B)^\bot\vdash A^\bot\oplus B^\bot$. Provided that I have correctly understood that **it is licit to commute (is it?)** any two formulae (or set of them) if the are either both before the $\vdash$ or both after it -I think that what my book calls *the two exchange rules* precisely are that- I would respectively prove the converse derivations $A^\bot\wp B^\bot\vdash (A\otimes B)^\bot$, $A^\bot\otimes B^\bot\vdash (A\wp B)^\bot$, $A^\bot\& B^\bot\vdash (A\oplus B)^\bot$ and $A^\bot\oplus B^\bot\vdash (A\& B)^\bot$ with the following calculations:

$$\dfrac{\dfrac{\dfrac{}{ A,A^\bot\vdash}\text{id}\quad\dfrac{}{B, B^\bot\vdash}\text{id}}{ A, B, A^\bot\wp B^\bot\vdash}\wp}{A^\bot\wp B^\bot, A\otimes B\vdash}\otimes \qquad \dfrac{\dfrac{\dfrac{}{A^\bot,A\vdash}\text{id}\quad\dfrac{}{B^\bot,B\vdash}\text{id}}{ A^\bot, B^\bot, A\wp B\vdash}\wp}{A\wp B, A^\bot\otimes B^\bot\vdash}\otimes$$ $$\dfrac{\dfrac{\dfrac{}{ A, A^\bot\vdash}\text{id}}{A, A^\bot\text{&}B^\bot \vdash}\text{&}\quad\dfrac{\dfrac{}{ B, B^\bot\vdash}\text{id}}{ B,A^\bot\text{&}B^\bot\vdash}\text{&}}{ A^\bot\text{&}B^\bot,A\oplus B\vdash}\oplus\qquad\dfrac{\dfrac{\dfrac{}{ A^\bot, A\vdash}\text{id}}{A^\bot, A\text{&}B \vdash}\text{&}\quad\dfrac{\dfrac{}{ B^\bot, B\vdash}\text{id}}{ B^\bot,A\text{&}B\vdash}\text{&}}{ A\text{&}B,A^\bot\oplus B^\bot\vdash}\oplus$$

**Are they correct?** I $\infty$-ly thank both the two people who already provided an answer and whoever will provide this remaining part!

***EDIT*** **Mar 22'15**: I decided to accept the first answer in order not to leave the question as unanswered, since I feel **almost** sure that it, together with my observations in the previous edit, solves my question, although the answerer cannot answer my doubts at present. I would be $\infty$-ly grateful to anybody **confirming** my calculcations' correctness and assumptions about commutativity **or explaining** where they are **wrong**.

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Following the request expressed in a comment by coffeemath, I am adding a brief informal description of the conjunctions and disjunctions, quoting from my book (D. Palladino, C. Palladino, *Logiche non classiche*):

$A\text{&}B$ *means that there is the possibility to chose only one of $A$ and* $B$;

$A\otimes B$ *means that both $A$ and $B$ can be available*;

$A\oplus B$ *means $A$ or $B$, but the choice of the conjunct is made by somebody else*;

$A\wp B$ *represents simultaneous targets*.