# Definition talk:Type

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I'm sorry, but I'm rolling back that last change:

For example, $\operatorname{tp}(b / A\cup\{b\}) = \operatorname{tp}(b / A,c) = \operatorname{tp}(b / Ac)$

back to

For example, $\operatorname{tp}(b / A\cup\{c\}) = \operatorname{tp}(b / A,c) = \operatorname{tp}(b / Ac)$

because the latter is consistent with the exposition. I confess I'm not too sure of my ground here, but the editor here is anonymous and I therefore may not easily be able to enter into conversation.

If User:qedetc is still around, I'd welcome a comment. --prime mover 00:16, 6 October 2011 (CDT)