What does $\bigwedge\bigwedge\bigvee$ mean in discrete mathematics?

Question

From Discrete Mathematics and its Applications, Kenneth H. Rosen, 7th Edition, Pg 54:

• For each cell with a given value, we assert p(i, j, n) when the cell in row I and column j has the given value n.
• We assert that every row contains every number: $\displaystyle\bigwedge\limits_{i=1}^9\bigwedge\limits_{n=1}^9\bigvee\limits_{j=1}^9p(i,j,n)$

I know that $\bigwedge$ and $\bigvee$ bear the same relation to $\land$ and $\lor$ respectively that $\sum$ does to +, i.e $\displaystyle\bigwedge\limits_{i=1}^np(i)$ can be interpreted as $p_1\land p_2\land p_3\land p_4...\land p_n$, but I don't know how to proceed when three symbols are together. Can someone please help :).

Answer 1

I will show for 2 indices instead of 9, as it is simpler to see. But firstly I want to say that in Mathematics often symbols are very very ugly, but once you decompose what each part means it's quite simple.

$$\displaystyle\bigwedge\limits_{i=1}^2\bigwedge\limits_{n=1}^2\bigvee\limits_{j=1}^2p(i,j,n)$$

Let us decompose this! Maybe in this form, it will already be obvious what this means.

$$ \bigwedge\limits_{i=1}^2\left(\bigwedge\limits_{n=1}^2\left( \bigvee\limits_{j=1}^2p(i,j,n) \right)\right) $$

If not, then let us decompose it, first we evaluate the inner parentheses:

$$ \bigwedge\limits_{i=1}^2\left(\bigwedge\limits_{n=1}^2[ p(i,1,n) \vee p(i,2,n) ]\right) $$

Then we evaluate the second parentheses:

$$ \bigwedge\limits_{i=1}^2 \Big( [ p(i,1,1) \vee p(i,2,1) ] \wedge [ p(i,1,2) \vee p(i,2,2) ] \Big) $$

Then finally we examine the last statement:

$$ \Big( [ p(1,1,1) \vee p(1,2,1) ] \wedge [ p(1,1,2) \vee p(1,2,2) ] \Big) \wedge \Big([ p(2,1,1) \vee p(2,2,1) ] \wedge [ p(2,1,2) \vee p(2,2,2) ] \Big) $$

If these expansions confuse you, a trick I used when I started with math several years ago is just imagine you have all the terms specified by the subscripts and superscripts, then 'combine' them with the operation given (the large symbol on the left, for example, $\Sigma$ means after having all the terms, we add all of them). Of course there are (not so) minor details like commutativity/ability to re-arrange terms or not etc.

Answer 2

Just resolve them one at a time. (I lowered the upper limits to 3 to make this shorter; if you keep them at $9$, you’ll end up with $9^3 = 729$ terms at the end.) \begin{align*} \bigwedge_{i=1}^3 \bigwedge_{n=1}^3 \bigvee_{j=1}^3 p(i,j,n) &= \bigwedge_{i=1}^3 \bigwedge_{n=1}^3 \big(p(i,1,n) \vee p(i,2,n) \vee p(i,3,n)\big) \\ &= \bigwedge_{i=1}^3 \Big(\big(p(i,1,1) \vee p(i,2,1) \vee p(i,3,1)\big) \\ &\qquad\qquad\wedge \big(p(i,1,2) \vee p(i,2,2) \vee p(i,3,2)\big)\\ &\qquad\qquad \wedge \big(p(i,1,3) \vee p(i,2,3) \vee p(i,3,3)\big) \Big) \\ &= \big(p(1,1,1) \vee p(1,2,1) \vee p(1,3,1)\big) \\ &\quad\wedge \big(p(1,1,2) \vee p(1,2,2) \vee p(1,3,2)\big) \\ &\quad\wedge \big(p(1,1,3) \vee p(1,2,3) \vee p(1,3,3)\big) \\ &\quad\wedge \big(p(2,1,1) \vee p(2,2,1) \vee p(2,3,1)\big) \\ &\quad\wedge \big(p(2,1,2) \vee p(2,2,2) \vee p(2,3,2)\big) \\ &\quad\wedge \big(p(2,1,3) \vee p(2,2,3) \vee p(2,3,3)\big) \\ &\quad\wedge \big(p(3,1,1) \vee p(3,2,1) \vee p(3,3,1)\big) \\ &\quad\wedge \big(p(3,1,2) \vee p(3,2,2) \vee p(3,3,2)\big) \\ &\quad\wedge \big(p(3,1,3) \vee p(3,2,3) \vee p(3,3,3)\big)\end{align*}

In practice, you might want to find a cleverer way to thing about these, of course, but that depends on what exactly $p(i,j,n)$ is.