Comma notation in universal qunatification

Is the expression $$\forall a,b,c \in M : \varphi(a,b,c)$$ equivalent to $$\forall a \forall b \forall c : (a \in M \land b \in M \land c \in M) \rightarrow \varphi(a,b,c)$$ ?

• Yes. That is so. – Graham Kemp Aug 1 at 6:58

The formula $$\forall a \in M : \varphi(a)$$ is syntactic sugar for $$\forall a \ (a \in M \to \varphi(a))$$ and $$\forall a, b, c \in M : \varphi(a,b,c)$$ is syntactic sugar for $$\forall a \in M \ \forall b \in M \ \forall c \in M : \varphi(a,b,c)$$ which according to the former syntactic sugar means $$\forall a \ (a \in M \to \forall b \ (b \in M \to \forall c \ (c \in M \to \varphi(a,b,c))))$$ which is logically equivalent to $$\forall a \ \forall b \ \forall c \ (a \in M \wedge b \in M \wedge c \in M \to \varphi(a,b,c))$$.
Yes, the former notation is an abbreviation of the latter. We often even leave out the commas, so you would see $$\forall abc \in M : \varphi(a,b,c)$$.
A similar abbreviation is used for the existential quantifier: $$\exists abc \in M : \varphi(a,b,c)$$ means $$\exists abc : a \in M \land b \in M \land c \in M \land \varphi(a, b, c).$$