Using $\bigvee$ and $\bigwedge$ instead $\exists$ and $\forall$

My professor of Algebra use some "strange" notation for me. He uses $\bigvee$ instead $\exists$ and $\bigwedge$ instead $\forall$. For example $$\displaystyle\bigwedge_{x\in \mathbb{Z}}\bigwedge_{m\in \mathbb{Z}\backslash\{0\}}\bigvee_{q,r\in \mathbb{Z}}(x=qm+r \wedge 0\leq r<|m|)$$ is same as $(\forall x \in \mathbb{Z})(\forall m \in \mathbb{Z}\backslash \{0\}) (\exists q,r \in \mathbb{Z}) (x=qm+r \wedge 0\leq r<|m|) $. If we know the set with which we are working, then we say $\displaystyle\bigwedge_{x}\bigvee_{y}(x+y=0)$ (without saying $x \in \text{Set}$). I asked him for this notation, and he said that I can see this in

K.Kuratowski, A.Mostowski, Set theory, PWN, Warszawa, 1976.

I found this book in library and it's really true.

Could someone say something more about this notation? Is this standard notation in mathematics? Did you see it anywhere else?

Solution 1:

I try to give you an argument why this notation makes sense.

Consider $$\displaystyle\bigvee_x A(x)$$ as an infinite version of $\vee$. For example, if $x$ comes from a countable set $\{x_1,x_2,x_3,\ldots\}$, then consider $\displaystyle\bigvee_x A(x)$ as $$A(x_1) \vee A(x_2) \vee A(x_3) \vee \ldots.$$ This expression is true as long as there is at least one $x_i$ such that $A(x_i)$ is true. So equivalently, there exists an $x_i$ such that $A(x_i)$ is true, which is $\displaystyle\exists x : A(x)$.

You can do the same for $\bigwedge$.