# Propositional logic notation conversions and naming.

Question 1

For example for: $$\bigvee_{i = 7}^{9} p_{i}$$

we write: $$\bigvee_{i = 7}^{9} p_{i} = p_{7}\vee p_{8}\vee p_{9}$$

What we write for? $$\bigwedge_{i=1}^{9} \bigwedge_{n=1}^{9} \bigvee_{j=1}^{9}~p(i,j,n)$$

Question 2

What is called that form?

$$\bigvee_{i = 7}^{9} p_{i}$$

And what is called that form?

$$p_{7}\vee p_{8}\vee p_{9}$$

$$\bigwedge_{i=1}^{2} \bigwedge_{n=1}^{2} \bigvee_{j=1}^{2}~p(i,j,n)=\\ =\left(\bigwedge_{n=1}^{2} \bigvee_{j=1}^{2}\ p(1,j,n)\right) \wedge \left(\bigwedge_{n=1}^{2} \bigvee_{j=1}^{2}\ p(2,j,n)\right)=\\ =\left[\left(\bigvee_{j=1}^{2}\ p(1,j,1)\wedge \bigvee_{j=1}^{2}\ p(1,j,2)\right)\right] \wedge \left[\left( \bigvee_{j=1}^{2}\ p(2,j,1)\wedge \bigvee_{j=1}^{2}\ p(2,j,2)\right)\right]=\\ =(p(1,1,1)\vee p(1,2,1))\wedge (p(1,1,2)\vee p(1,2,2))\wedge (p(2,1,1)\vee p(2,2,1))\wedge (p(2,1,2)\vee p(2,2,2))$$

This is the compact form: $$\bigvee_{i=7}^{9}p_i$$

This is the expanded form: $$p_7\vee p_8\vee p_9$$

• If we remove [ ] it will mean the same, right? Are they for just better visual separation of the parts? – vasili111 Sep 15 '18 at 11:52
• yes, indeed, they can be removed. – trying Oct 7 '18 at 15:15

$\bigvee_{i = 7}^{9} p_{i}$ and $p_{7}\vee p_{8}\vee p_{9}$ are just two different notations for the same formula (they denote the same object). So, if your question refers to notations, $\bigvee_{i = 7}^{9} p_{i}$ is the compact or implicit notation and $p_{7}\vee p_{8}\vee p_{9}$ is the expanded or explicit notation for such a formula. But if your question refers to the (same) formula denoted by $\bigvee_{i = 7}^{9} p_{i}$ and $p_{7}\vee p_{8}\vee p_{9}$, such a formula is a (disjunctive) clause, i.e. a disjunction of literals, where a literal is an atomic formula or its negation. Since it is a disjunctive clause, such a formula is both a conjunctive normal form and a disjunctive normal form.
• @vasili111 - No, they aren't. If you replace "9" with "2", then $\bigwedge_{i=1}^{2} \bigvee_{j=1}^{2} \bigwedge_{n=1}^{2}~p_{i,j,n} = \bigwedge_{i=1}^{2} \bigvee_{j=1}^{2} (p_{i,j,1} \land p_{i,j,2}) = \bigwedge_{i=1}^{2} \big( (p_{i,1,1} \land p_{i,1,2}) \lor (p_{i,2,1} \land p_{i,2,2}) \big) = \big( (p_{1,1,1} \land p_{1,1,2}) \lor (p_{1,2,1} \land p_{1,2,2}) \big) \land \big( (p_{2,1,1} \land p_{2,1,2}) \lor (p_{2,2,1} \land p_{2,2,2}) \big)$; – Taroccoesbrocco Aug 24 '18 at 14:58
• @vasili111 - whereas $\bigwedge_{i=1}^{2} \bigwedge_{n=1}^{2} \bigvee_{j=1}^{2}~p_{i,j,n} = \bigwedge_{i=1}^{2} \bigwedge_{n=1}^{2} (p_{i,1,n} \lor p_{i,2,n}) = \bigwedge_{i=1}^{2} \big((p_{i,1,1} \lor p_{i,2,1}) \land (p_{i,1,2} \lor p_{i,2,2}) \big) = \big((p_{1,1,1} \lor p_{1,2,1}) \land (p_{1,1,2} \lor p_{1,2,2}) \big) \land \big((p_{2,1,1} \lor p_{2,2,1}) \land (p_{2,1,2} \lor p_{2,2,2}) \big)$. – Taroccoesbrocco Aug 24 '18 at 15:00