Precedence of Operations and Negating Compound Propositions

Precedence

The precedence of the negation operation is the highest after the parenthesis, and it is resolved before any other operator. For example, in $\neg p \lor q$, the disjunction operation will always take $\neg p$ as an operand unless parenthesized otherwise. This means:

$$\neg p \lor q \equiv \left(\neg p\right) \lor q.$$

This is clearer if we use $\overline p$ to denote the negation of $p$. In this notation,

$${\overline p} \lor q \equiv ({\overline p}) \lor q.$$

It is important to observe that,

$$\overline{ \left(p \lor q\right)} \equiv \neg \left(p \lor q\right) \not \equiv \left(\neg p\right) \lor q \equiv ({\overline p}) \lor q.$$

The table below gives the order of precedence among conjunction, disjunction, negation, and parenthesization. Note that the higher the number, the lower its precedence.

Operation	Symbol	Precedence (Order of application)
Parenthesis	$( \ )$	1
Negation	$\neg$	2
Conjunction	$\land$	3
Disjunction	$\lor$	4

Just like BODMAS helps us remember the precedences of arithmetic operators, we can use the acronym PNCD to remember the precedences of logical operators. PNCD stands for parentheses, negation, conjunction, and disjunction.

Examples

If we apply the operations in the correct order of precedence, we can write $G_1$ as:

$$(p_b \land p_h) \lor (q_b \land q_h).$$

This can be translated into English as:

• $G_1 :$ Paul plays badminton and hockey or Quincy plays badminton and hockey.

A clearer way of saying this in everyday language is:

Either Paul or Quincy play both sports, hockey and badminton.

If we insist on applying the disjunction between $p_h$ and $q_b$ first, we must do that by explicitly enforcing it through parenthesis. Consider,

$$G_2 \equiv p_b \land \left(p_h \lor q_b \right) \land q_h.$$

This is different from $G_1$. Let’s use the commutativity of conjunction and write it as:

$$G_2 \equiv p_b \land q_h \land \left(p_h \lor q_b \right).$$

Now, we can translate it into everyday language quite easily:

• $G_2 :$ Paul plays badminton, Quincy plays hockey, and either Paul plays hockey, or Quincy plays badminton.

Why are $G_1$ and $G_2$ different? Consider the case when Paul plays both sports and Quincy plays none. In this case, $G_1$ is true. However, $G_2$ is false.

Let’s look at two propositions:

• $p$: Paul is an actor.

• $q$: Quentin is a director.

Consider the compound proposition:

$$C_1:\neg (p \land q).$$

We can write it in everyday language as,

• $C_1$: It is not the case that Paul is an actor and Quentin is a director.

Let’s also make another compound proposition:

$$C_2:\neg p \land \neg q .$$

Let’s carefully make a truth table of these two compound statements:

$p$	$q$	$\neg p$	$\neg q$	$\neg (p \land q)$	$\neg p \land \neg q$
T	T	F	F	F	F
T	F	F	T	T	F
F	T	T	F	T	F
F	F	T	T	T	T

Voilà! We realize that the truth table of these two propositions is entirely different.

Let’s write $C_2$ in everyday language:

• $C_2:$ It is not the case that Paul is an actor, and it is not the case that Quentin is a director.

We simplify this as follows:

• $C_2:$ Neither Paul is an actor nor Quentin is a director.

Consider the case when exactly one of the statements among $p$ and $q$ is true and the other false. In this case, $C_1$ is true. However, $C_2$ is false.

We realize that $C_1$ and $C_2$ are completely different statements.

Let’s see the situation when we negate a disjunction.

We must be cautious while dealing with the negation operator; it is not an innocent operator. It is clear from the example above that:

$$\neg\left(p \land q\right) \not\equiv \neg p \land \neg q.$$

Similarly, the quiz must have convinced you that,

$$\neg\left(p \lor q\right) \not\equiv \neg p \lor \neg q.$$

The topic of the next lesson is how to negate a compound statement that is a conjunction or a disjunction of two statements.

Quiz