# 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?

• I suppose the motivation behind is something of this sort: $$(\forall x\in \{a,b,c\})(P(x))\iff P(a)\wedge P(b) \wedge P(c)$$ and $$(\exists x\in \{a,b,c\})(P(x))\iff P(a)\vee P(b)\vee P(c)$$ – Git Gud Mar 21 '13 at 18:22
• The notation is very much a minority notation. It is best, I think, to use notations of the majority. – André Nicolas Mar 21 '13 at 18:47
• I think this notation is very non-standard. I saw it being used only in texts on Infinitary Logic (plato.stanford.edu/entries/logic-infinitary ). – Yury Mar 21 '13 at 18:48
• @Yury: Inifinitary logic would use this notation because it admits formulae of infinite length (so there could be inifnitely many disjuncts/conjuncts in a formula). In that context, the notation would be very standard. In the context the OP is speaking of, it is quite nonstandard. Though it does hint at some of the reasoning behind the intuitionists' denial of the Law of Excluded Middle: taking a quantifier to be an abbreviation of an infinitely long sentence of this type, one cannot rely on one's intuition about finite sentences to conclude that either it or its negation must be true. – user642796 Mar 21 '13 at 18:55
• I have seen it in a number of places, but it’s not common. – Brian M. Scott Mar 21 '13 at 18:55

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 $\bigwedge$.
• The notation is by far not as common as $\exists$ and $\forall$, but still you can find it every now and then. – azimut Mar 21 '13 at 18:43