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

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  • $\begingroup$ Yes. That is so. $\endgroup$ – Graham Kemp Aug 1 at 6:58
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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 $\forall a \ \forall b \ \forall c \ (a \in M \wedge b \in M \wedge c \in M \to \varphi(a,b,c))$.

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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 $$ \exists abc : a \in M \land b \in M \land c \in M \land \varphi(a, b, c). $$

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