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 . Tijfo098 (talk) 08:43, 16 April 2011 (UTC)

Diagram

Someone may want to add here the diagram from 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)