Relation between consistency of ZF, MK and NBG What is relation between their consistency?
Is MK consistent implies NBG consistent  implies ZF consistent?
Or MK consistent implies NBG consistent. And MK consistent implies ZF consistent. NBG consistent implies ZF consistent. And there is no relation between consistency of MK and NBG.
Or is there no relation at all?
 A: ZF is consistent iff NBG is consistent.  If MK is consistent, then ZF and NBG are consistent, but this cannot be reversed (working in MK, say).  So the relative consistency strengths are $MK>NBG=ZF$.
Some parts of this are trivial.  MK has strictly more axioms than NBG, so if MK is consistent then trivially NBG is consistent.  Similarly, NBG includes all of the ZF axioms when you look only at sets, so if NBG is consistent then trivally ZF is consistent.
Somewhat less trivial is that if ZF is consistent, then NBG is consistent.  You can prove this by proving that if you have any model $(M,\epsilon)$ of ZF, then taking all definable subsets of the structure $(M,\epsilon)$ as your classes, you get a model of NBG. (Here I assume you do not include any choice axiom in NBG; if you include global choice, then you should first replace $M$ with its constructible universe $L^M$.)
Also less trivial is that consistency of MK is strictly stronger than that of ZF or NBG.  This follows from the fact that MK proves ZF is consistent (basically, it is capable of formalizing the argument "$V$ is a model of ZF and therefore ZF is consistent").  Since MK cannot prove its own consistency by Gödel (assuming it is consistent), it cannot prove that if ZF is consistent, then MK is consistent.
