Which propositional logics do you get by taking the class of Boolean algebras as a semantics and adding strong negation?
I've encountered weak negation in unrelated contexts before as a unary connective that exchanges designated and undesignated truth values, so I'm curious what happens if you staple it to a well-known algebraic semantics.
Let $\mathsf{I}(\varphi)$ denote the interpretation of $\varphi$. Let $X$ be a set and $\Lambda$ be the partially ordered set $(2^X, \subset)$.
- $ \mathsf{I}(a)$ is the interpretation of $a$ when $a$ is a propositional variable.
- $ \mathsf{I}(a \land b)$ is $\mathsf{I}(a) \cap \mathsf{I}(b)$.
- $ \mathsf{I}(a \lor b)$ is $\mathsf{I}(a) \cup \mathsf{I}(b)$.
- $\mathsf{I}(\lnot a)$ is $X \setminus \mathsf{I}(a)$. This is ordinary negation.
- $\mathsf{I}(!a)$ is $\varnothing$ if $\mathsf{I}(a)$ is $X$, otherwise $\mathsf{I}(a)$ is $X$. This is weak negation. It exchanges designated and undesignated truth values.
If $X$ is $\{e\}$, then this semantics reduces to classical logic and $\lnot$ and $!$ are indistinguishable.
By a theorem cited in this answer and discussed more generally in this article, the semantics obtained using the class of Boolean algebras is the same as the one obtained from using $\{\varnothing, \{e\}\}$ alone.
However, adding weak negation essentially on a whim is enough to make the resulting semantics non-classical.
The following is a theorem of classical logic because the two connectives have the same meaning in that setting, and a theorem in the Boolean algebra semantics because $\mathsf{I}(a)$ must be $\varnothing$ when $\mathsf{\lnot a}$ is $X$.
$$ \frac{\lnot a}{! a} $$
However, the following is a theorem of classical logic only.
$$ \frac{! a}{\lnot a} $$
The following inference rule also fails in the $!$-setting.
$$ \frac{a \lor b \;\;\text{and}\;\; !a}{b} \;\; \text{is not a valid inference rule} $$