Non-commutative Logical Operations My question is quite simple: Are there any logics in which the AND or NOR operations are non-commutative? If so, which ones and how do they operate?
 A: This is an ill-defined question. What properties of a logical connective warrant it being called AND or NOR which exclude commutativity? Why can't I take classical propositional logic and simply call implication, say, conjunction and say that's an answer to your question?
One way to meaningfully answer this question is to look at the rules of inference that define conjunction, see where commutativity comes from, and see if we can relax some properties while still producing something that could morally be called a form of conjunction. (The tl;dr version is: examples are ordered linear logic and other non-commutative substructural logics.)
Using a sequent calculus presentation, the rules for conjunction are as follows: $$\cfrac{\Gamma,A,B,\Delta\vdash C}{\Gamma,A\land B,\Delta\vdash C}(\land L) \qquad \cfrac{\Gamma\vdash A \quad \Delta\vdash B}{\Gamma,\Delta \vdash A\land B}(\land R)$$ There are many variations on how to present the rules for conjunction even within the sequent calculus. If you look at the sequent calculus page, you'll see that this presentation corresponds to neither of the sets of rules given for conjunction. For example, in LJ, the intuitionistic variant of LK as described on that page, the rules are given as: $$\cfrac{\Gamma,A,\Delta\vdash C}{\Gamma,A\land B,\Delta\vdash C}\quad\cfrac{\Gamma,B,\Delta\vdash C}{\Gamma,A\land B,\Delta\vdash C}\qquad\cfrac{\Gamma\vdash A\quad \Gamma\vdash B}{\Gamma\vdash A\land B}$$ These variations will turn out to be not so inconsequential for us.
To explain the notation a bit, $\Gamma \vdash A$ means the context $\Gamma$ entails the proposition $A$ where a context is a list of propositions.  The rules above are actually part of a definition of what entailment means. I write $\Gamma,\Delta$, where $\Delta$ is another context, to mean their concatenation with $\Gamma,A$ treating the proposition $A$ as a single element list.  For our purposes, the rules above can be thought of as metalogical implications.  For example, the $\land R$ rule means "if $\Gamma\vdash A$ and $\Delta\vdash B$ then $\Gamma,\Delta\vdash A\land B$" where the "if-then" and "and" are happening metalogically.
So why is $\land$ commutative? It's commutative because we can swap $A$ and $B$ in the conclusions of the rules above getting an entailment of $B\land A$ and still derive the conclusion. Why can we do that? The above rules alone certainly don't imply that. The key is an additional rule called exchange: $$\cfrac{\Gamma,A,B,\Delta\vdash C}{\Gamma,B,A,\Delta\vdash C}(Exchange)$$
This is an example of a structural rule. By applying this rule repeatedly, we can arbitrarily permute the list of propositions in a context.  With this rule, we can think of the context as a (finite) multiset or bag of propositions rather than a list. An additional structural rule, contraction, allows us to combine duplicate propositions which further allows us to think of a context as a (finite) set of propositions. A third rule, weakening, allows us to add extraneous propositions to the context of any entailment.  Substructural logics do away with some or all of these rules. If we drop all of these rules we get ordered linear logic.  (One can go further and not assume the context is a list, i.e. that concatenation of contexts is associative. This leads to bunched logic.)
With the structural rules, you can show that the two sets of rules I gave above for $\land$ are equivalent, i.e. you can derive one set from the other.  Without the structural rules, i.e. in ordered linear logic (or even linear logic which keeps the exchange rule), they split into two separate notions.  The first leads to multiplicative conjunction and the second to additive conjunction.  In linear logic, these will still be commutative. In ordered linear logic, which doesn't have the exchange rule, multiplicative conjunction further splits into two notions: left multiplicative conjunction and right multiplicative conjunction.  The first set of rules I gave above then corresponds to left multiplicative conjunction. This splitting of conjunction into multiple distinct concepts happens for the other logical connectives as well.
A common interpretation of linear logic (with exchange) is as a logic of resource consumption. If a snack costs \$1 and a drink costs \$1 and I have \$1, then I can't end up with both a snack and a drink. In linear logic, this could look like the fact that I can't derive $$\cfrac{\$1\vdash\mathsf{snack}\quad\$1\vdash\mathsf{drink}}{\$1\vdash\mathsf{snack}\otimes\mathsf{drink}}$$ where here I'm writing $\otimes$ for the multiplicative conjunction as is common in linear logic. Ordered linear logic arose from Lambek's work in linguistics where the context would be the sequence of words in a sentence where order clearly matters, particularly for positional languages like English. However, I'll give a more computational interpretation. We can think of $\Gamma\vdash A$ as meaning "after processing the messages in $\Gamma$ in left-to-right order, I'll produce the message $A$". Renaming $\land$ in $\land L$ and $\land R$ above to $\vartriangleright$, then ${\vartriangleright}L$ means "processing the message $A\vartriangleright B$ is the same as processing the message $A$ followed by the message $B$". ${\vartriangleright}R$ means "if after processing $\Gamma$ I'd produce $A$ and after processing $\Delta$ I'd produce $B$, then after processing $\Gamma$ followed by $\Delta$ I'd produce $A\vartriangleright B$" which is to say I'd produce $A$ followed by $B$.
A: An operator '$\operatorname{op}$' in Boolean algebra is commutative iff:
$$A \operatorname{op} B=B \operatorname{op} A$$
which is equivalent to testing if:
$$0 \operatorname{op} 1=1 \operatorname{op} 0$$
Both AND or NOR pass this, and are commutative.
A: Sure. Short circuiting AND & OR in most programming languages are noncommutative, if one of the included expressions can halt (i.e. get stuck in an infinite loop)
