Talk:Leibniz operator

Mistake?

The article states:

$\phi \leftrightarrow \psi \in T$

that defines $\phi \equiv _{T}\psi$ is equivalent to the condition

$T\vdash _{\mathcal {S}}\phi$ if and only if $T\vdash _{\mathcal {S}}\psi$.

This is only true if T is a complete theory. Is it possible the article means :$T\vdash _{\mathcal {S}}\chi [\phi ]$ if and only if $T\vdash _{\mathcal {S}}\chi [\psi ]$?

Why Leibniz ?

It would be nice to see some indication of the rationale for this operator being named for Leibniz; what part of his work (presumably in a precursor of algebraic logic) does it encapsulate ? (c.f. my rationale for applying the same name to tensor operators obeying Leibniz's product rule, which I had done before hearing of the name's use in algebraic logic.) 84.215.6.188 (talk) 15:22, 4 January 2011 (UTC)

I can't tell for sure, but I think it is a reference to Leibniz's law.—Emil J. 15:30, 4 January 2011 (UTC)

Yes [1]. Tijfo098 (talk) 08:43, 16 April 2011 (UTC)

Diagram

Someone may want to add here the diagram from [2] p. 25. Of course, it would be better if all those notions are defined in the wiki article, which is currently far from doing, even for those classes that it does mention. Tijfo098 (talk) 05:57, 13 April 2011 (UTC)