> There is indeed an implication operator on partitions.

>

> [David Ellerman](https://arxiv.org/pdf/0902.1950.pdf) defines it as such:

>

> \\[ [[x] \Rightarrow [y]] : = \mathrm{int}([x]^c \cup [y]) \\]

(I changed your post to reflect Ellerman's original notation)

While Ellerman does define a conditional that way, \\((x \wedge - )\\) does not appear to be the left adjoint of \\((x \Rightarrow - )\\).

There are other conditionals in logic that are not right adjoints of \\(x \wedge - \\). One example is David Lewis' [counterfactual conditional](https://plato.stanford.edu/entries/causation-counterfactual/#CouCauDep). Other examples include also Gillies' [*indicative conditionals*](http://rci.rutgers.edu/~thony/resources/phil-language-and-indicatives-radio-edit.pdf) and the *probability conditional* (see [Adams 1998, pg. 154](http://press.uchicago.edu/ucp/books/book/distributed/P/bo3636904.html)).

We don't need all of the adjoint functor theorem for posets to see that there's no right adjoint for \\(x \wedge - \\) for partitions.

Let's just review the relevant part of the adjoint functor theorem for posets to see why it's not going to work.

Let \\(f: X \to X\\) and \\(g : X \to X\\) by monotone endofunctors on a lattice \\((X, \le, \wedge, \vee)\\), such that \\(g \dashv f\\).

We have:

\[

\begin{align}

g(x \vee y) \le z & \iff x \vee y \le f(z) \\\\

& \iff x \le f(z) \text{ and } y \le f(z) \\\\

& \iff g(x) \le z \text{ and } g(y) \le z \\\\

& \iff g(x) \vee g(y) \le z

\end{align}

\]

Thus \\(g(x \vee y) \le z \iff g(x) \vee g(y) \le z\\).

But then, since \\(g(x \vee y) \le g(x \vee y) \\) we have \\(g(x) \vee g(y) \le g(x \vee y)\\).

Dually, since \\(g(x) \vee g(y) \le g(x) \vee g(y)\\) we have \\(g(x \vee y) \le g(x) \vee g(y)\\).

Hence \\(g(x \vee y) = g(x) \vee g(y)\\).

Now if you look at [#15](https://forum.azimuthproject.org/discussion/comment/20139/#Comment_20139) I give an example of where \\(A \wedge (B \vee C) \neq (A \wedge B) \vee (A \wedge C)\\).

So Ellerman's conditional must break one of the usual rules (perhaps *modus ponens* or the *hypothetical syllogism* that Jonathan pointed out).

>

> [David Ellerman](https://arxiv.org/pdf/0902.1950.pdf) defines it as such:

>

> \\[ [[x] \Rightarrow [y]] : = \mathrm{int}([x]^c \cup [y]) \\]

(I changed your post to reflect Ellerman's original notation)

While Ellerman does define a conditional that way, \\((x \wedge - )\\) does not appear to be the left adjoint of \\((x \Rightarrow - )\\).

There are other conditionals in logic that are not right adjoints of \\(x \wedge - \\). One example is David Lewis' [counterfactual conditional](https://plato.stanford.edu/entries/causation-counterfactual/#CouCauDep). Other examples include also Gillies' [*indicative conditionals*](http://rci.rutgers.edu/~thony/resources/phil-language-and-indicatives-radio-edit.pdf) and the *probability conditional* (see [Adams 1998, pg. 154](http://press.uchicago.edu/ucp/books/book/distributed/P/bo3636904.html)).

We don't need all of the adjoint functor theorem for posets to see that there's no right adjoint for \\(x \wedge - \\) for partitions.

Let's just review the relevant part of the adjoint functor theorem for posets to see why it's not going to work.

Let \\(f: X \to X\\) and \\(g : X \to X\\) by monotone endofunctors on a lattice \\((X, \le, \wedge, \vee)\\), such that \\(g \dashv f\\).

We have:

\[

\begin{align}

g(x \vee y) \le z & \iff x \vee y \le f(z) \\\\

& \iff x \le f(z) \text{ and } y \le f(z) \\\\

& \iff g(x) \le z \text{ and } g(y) \le z \\\\

& \iff g(x) \vee g(y) \le z

\end{align}

\]

Thus \\(g(x \vee y) \le z \iff g(x) \vee g(y) \le z\\).

But then, since \\(g(x \vee y) \le g(x \vee y) \\) we have \\(g(x) \vee g(y) \le g(x \vee y)\\).

Dually, since \\(g(x) \vee g(y) \le g(x) \vee g(y)\\) we have \\(g(x \vee y) \le g(x) \vee g(y)\\).

Hence \\(g(x \vee y) = g(x) \vee g(y)\\).

Now if you look at [#15](https://forum.azimuthproject.org/discussion/comment/20139/#Comment_20139) I give an example of where \\(A \wedge (B \vee C) \neq (A \wedge B) \vee (A \wedge C)\\).

So Ellerman's conditional must break one of the usual rules (perhaps *modus ponens* or the *hypothetical syllogism* that Jonathan pointed out).