Semantic/Syntactic Monotonicity 
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*Monotonicity is defined as the implication: M $\vdash$ A $\to$ M $ \cup$ N $\vdash$ A. It depends on the calculus if it is true or not. Correct?


*But the following semantic version is always true in classical logic: M $\vDash$ A $\to$ M $ \cup$ N $\vDash$ A. Correct?


*What is the name of the property of 2. because it seems to me that monotonicity is only used as a (syntactical) proof concept as in 1.


*How do you prove 1. for a proof system P since you cannot prove it in P itself because it is metalanguage? How does it work conceptually?


*How do you prove 2.?
 A: 

*

*Monotonicity is defined as the implication: M $\vdash$ A $\to$ M $ \cup$ N $\vdash$ A. It depends on the calculus if it is true or not. Correct?


Yes, but it will hold for any calculus that is complete for classical logic.



*But the following semantic version is always true in classical logic: M $\vDash$ A $\to$ M $ \cup$ N $\vDash$ A. Correct?


Yes.



*What is the name of the property of 2. because it seems to me that monotonicity is only used as a (syntactical) proof concept as in 1.


If anything I'd say that monotonicity is a more semantic notion, and a term that is more closely associated with proof systems is weakening.



*How do you prove 1. for a proof system P since you cannot prove it in P itself because it is metalanguage? How does it work conceptually?


Yes, you will have to make a proof arguing about the calculus, though usually it's pretty much a direct consequence of how the proof system is set up. E.g. in natural deduction, $M \vdash A$ is defined to hold iff there exists a derivation with all the right rule applications ending in $A$ and the open assumptions a subset of $M$. Now of course every subset of $M$ is also a subset of $M \cup N$, and so the same derivation serves as a proof of $M \cup N \vdash A$.



*How do you prove 2.?


See Mauro Allegranza's answer: Every model of $M \cup N$ is also a model of $M$ and therefore, by assumption, a model of $A$, hence $M \cup N \vDash A$.
