Bi-implication

What is a bi-implication?

As $p \Rightarrow q$ is not logically equivalent to $q \Rightarrow p$, it is interesting to look at the following proposition:

$$p \Leftrightarrow q:= \left(p \Rightarrow q\right) \land \left(q \Rightarrow p\right).$$

Note: The $:=$ symbol used above means “defined as” or “is by definition equal to.”

We call $p \Leftrightarrow q$ a bi-implication. It is also called a biconditional operation. In English, we can state it in different ways. A few of the commonly used phrases are as follows:

• $q$ if $p$ and $p$ if $q$.
• $p$ if and only if $q$.
• $p$ iff $q$. (Here, iff is a short form of “if and only if.”)
• $p$ implies $q$ and converse.
• $p$ is necessary and sufficient for $q$.

Truth table

Let’s look at the truth table of bi-implication. Note that bi-implication is different from simple implication and its converse.

$p$	$q$	$p \Rightarrow q$	$q \Rightarrow p$	$p \Leftrightarrow q$
T	T	T	T	T
T	F	F	T	F
F	T	T	F	F
F	F	T	T	T

The bi-implication is true if both operands have the same truth value. We can also use this information in the following way.

Note: If bi-implication is true, we can conclude that both of its operands are logically equivalent.

This way, the bi-implication allows us to establish that two propositions are equivalent. If we want to prove that:

$$q_1 \equiv q_2$$

we can achieve it by showing that,

• $q_1 \Rightarrow q_2$ is true,

and

• $q_2 \Rightarrow q_1$ is also true,

hence establishing that,

• $q_1 \Leftrightarrow q_2$ is true.

Properties

As $p \Leftrightarrow q$ is only true when both $p$ and $q$ have the same truth value, it is easy to conclude that,

$$p \Leftrightarrow q \equiv q \Leftrightarrow p.$$

That is, bi-implication is commutative. Let’s look at the following truth table for the associative property. Assume three arbitrary propositions $p$, $q$, and $r$. The truth value of all three is shown in the first column collectively.

$\left(p, q, r\right)$	$p \Leftrightarrow q$	$q \Leftrightarrow r$	$\left(p \Leftrightarrow q\right) \Leftrightarrow r$	$p \Leftrightarrow \left(q \Leftrightarrow r\right)$
(T,T,T)	T	T	T	T
(T,T,F)	T	F	F	F
(T,F,T)	F	F	F	F
(T,F,F)	F	T	T	T
(F,T,T)	F	T	F	F
(F,T,F)	F	F	T	T
(F,F,T)	T	F	T	T
(F,F,F)	T	T	F	F

If we compare the last two columns of the truth table, associativity is evident. That is;

$$\left(p \Leftrightarrow q\right) \Leftrightarrow r \equiv p \Leftrightarrow \left(q \Leftrightarrow r\right).$$

Note: Bi-implication is commutative and associative.

Examples

Let’s look at an example that will help us understand the difference between implication and bi-implication.

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Consider the following propositions:

• $q_1$: Isra scored more than 80% marks in the final examination of the algebra class.
• $q_2$: Isra got an A grade in algebra.

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Now,

• $p_1=q_1 \Rightarrow q_2$: If Isra scored more than 80% marks in the final examination of algebra then she got an A grade in algebra.
• $p_2=q_2 \Rightarrow q_1$: If Isra got an A grade in algebra then she scored more than 80% marks in the final examination of algebra.
• $p_3=q_1 \Leftrightarrow q_2$: Isra scored more than 80% marks in the final examination of algebra if and only if she got an A grade in algebra.

There are four possible scenarios, which are depicted in the following truth table:

$q_1$	$q_2$	$p_1$	$p_2$	$p_3$
T	T	T	T	T
T	F	F	T	F
F	T	T	F	F
F	F	T	T	T

Let’s look at the table closely and discuss each scenario that corresponds to a particular row of the truth table.

Scenario 1:

Assume that Isra scored more than 80% marks in the final examination of the algebra class; that is, $q_1$ is true, and she got an A grade in algebra; that is, $q_2$ is also true. This scenario corresponds to the first row of the truth table. Because hypothesis and conclusion both are true for $p_1$ and $p_2$, they are true. Because both $p_1$ and $p_2$ are true therefore, $p_3$ is also true.

Scenario 2:

Assume that Isra scored more than 80% marks in the final examination of algebra; that is, $q_1$ is true, and she did not get an A grade in algebra; that is, $q_2$ is false. This scenario corresponds to the second row of the truth table. For $p_1$, the hypothesis is true, and the conclusion is false; therefore, it is false. For $p_2$, the hypothesis is false; therefore, it is true. Because $p_1$ and $p_2$ have different truth values; therefore, $p_3$ is false.

Scenario 3:

Assume that Isra did not score more than 80% marks in the final examination of algebra; that is, $q_1$ is false, and she got an A grade in algebra; that is, $q_2$ is true. This scenario corresponds to the third row of the truth table. For $p_1$, the hypothesis is false. Therefore, it is true. But for $p_2$, the hypothesis is true, and the conclusion is false, making it false. Because both $p_1$ and $p_2$ have different truth values, therefore, $p_3$ is false.

Scenario 4:

Assume that Isra did not score more than 80% marks in the final examination of algebra; that is, $q_1$ is false, and she did not get an A grade in algebra; that is, $q_2$ is also false. This scenario corresponds to the fourth row of the truth table. Because the hypotheses are false for both $p_1$ and $p_2$; hence, they are true. Therefore, because both $p_1$ and $p_2$ are true, $p_3$ is also true.

Let’s look at another example.

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Consider:

• $p_4$: Deniz will open his umbrella if and only if it is raining.

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The statement $p_4$ is saying that if Deniz will open his umbrella, then it is raining and if it is raining, then Deniz will open his umbrella.

When true?

The proposition $p_4$ is true if Deniz opens up his umbrella whenever it rains, and whenever Deniz opens up his umbrella, it is raining.

When false?

The proposition $p_4$ is false if Deniz opens up his umbrella when it is not raining or when Deniz does not open up his umbrella during rain.