Iff Interpretation

Question

I understand that

(1) "$A$ if and only if $B$" ($A\iff B)$

means that

(2) "$A$ implies $B$ and $B$ implies $A$" $(A\implies B)\land (B\implies A)$.

The phrase "$A$ if and only if $B$" sounds as though something applies exclusively or precisely in case something else is true, i.e., not only does $B$ make $A$ true, but $B$ is the only thing that makes $A$ true. If this is the case, then how exactly does this interpretation relate to two things implying one another?

For the Bounty:
(1) I think that a natural interpretation of the statement $A$ if and only if $B$ is that $A$ is true precisely in case $B$ is true, i.e., if $B$ is true, then $A$ is true, and if $B$ is false, then $A$ is false. Is this correct? And if so, how does this interpretation mean that $A$ and $B$ imply one another and also have the same truth value?

(2) Related to (1), the (true) statement $\forall x\in\mathbb R:2x=5\iff x={5\over 2}$ means that for any replacement of the variable, the two statements have the same truth value. How does two statements having the same truth value in all cases mean that they imply each other?

(3) Most importantly, I need exposition. Am I missing anything obvious, or is this a simple thing?

Answer 1

Definition 0. By a boolean, I mean an element of $\{0,1\}$. We think of $1$ as "true" and $0$ as "false."

Definition 1. Think of "and" as a function that accepts a pair of booleans $(a,b)$, and returns a third boolean, denoted $a \wedge b$. This function is defined as follows:

For all $a,b \in \{0,1\}$:

• If $a=1$ and $b=1$, then $a \wedge b = 1$.
• If $a\neq 1$ or $b\neq 1$, then $a \wedge b = 0$.

We can use the above definition to check the following:

$$0 \wedge 0 = 0, \quad 1 \wedge 1 = 1,\quad 1 \wedge 0 = 0, \quad 0 \wedge 1 = 0$$

This information can be neatly arranged into a $2 \times 2$ table (try it!).

Definition 1. We think of "if and only if" as a function that accepts a pair of booleans $(a,b)$, and returns a third boolean, denoted $a \iff b$. This function is defined as follows:

For all $a,b \in \{0,1\}$:

• If $a=b$, then $a \iff b = 1$.
• If $a \neq b$, then $a \iff b = 0$.

$$0 \iff 0 = 1, \quad 1 \iff 1 = 1,\quad 1 \iff 0 = 0, \quad 0 \iff 1 = 0$$ Once again, try putting this into a $2 \times 2$ table.

Definition 2. We think of "implies" as a function that accepts a pair of booleans $(a,b)$, and returns a third boolean, denoted $a \implies b$. This function is defined as follows:

For all $a,b \in \{0,1\}$:

• If $a=1$ and $b=0$, then $a \implies b = 0$.
• If $a \neq 1$ or $b \neq 0$, then $a \implies b = 1$.

$$0 \implies 0 = 1, \quad 1 \implies 1 = 1,\quad 1 \implies 0 = 0, \quad 0 \implies 1 = 1$$

Table? Good.

We're now ready to prove:

Proposition. For all booleans $a$ and $b$, we have:

$$(a \iff b) = (a \implies b) \wedge (b \implies a)$$

To prove this, we have to check $4$ cases:

1. $a=0,b=0$
2. $a=1,b=1$
3. $a=1,b=0$
4. $a=0,b=1$

I'll do Case (2) for you:

• $\mathrm{LHS} = (1 \iff 1) = 1$

• $\mathrm{RHS} = (1 \implies 1) \wedge (1 \implies 1) = 1 \wedge 1 = 1$

So $\mathrm{LHS} = \mathrm{RHS}$

Now you just have to check the other three cases.

Answer 2

Let's assume that $A$ is true only if $B$ is true. Now if we know that $A$ is true, $B$ must also be true, because we assumed that then only can $A$ be true. This means we could deduce that $B$ is true from the fact that $A$ was true. In other words, the truth of $A$ implies the truth of $B$, denoted $ A \implies B$.

Now lets assume that $A$ is true if $B$ is true. If we know that $B$ is true then we can deduce that $A$ is also true under this assumption. This means that $B$ implies $A$ denoted $B \implies A$

So if we assume both these things ($A$ is true if and only if $B$ is true), we have that A implies B and B implies A, denoted $A \iff B$

Answer 3

"A if and only if B" can be separated into two statements: A if B, and A only if B. "A if and only if B" says that both of these are true.

"A if B" means $B\implies A$. In other words, if B is true then so is A. For certain statements A and B, A will be true at least some of the time when B is false.

"A only if B" means A is only true if B is true. In other words, if A is true then so is B. For certain statements A and B, B will be true at least some of the time when A is false. Thus "A only if B" means $A\implies B$.

Answer 4

Let $A $ if and only if $B$. This means $B \implies A$, since $A $ if $B$ is given.

How you interpret the only part is what is interesting. $A$ only if $B$ seems to means something along the lines of : "If C is any proposition not identically equal to B, then C doesn't imply A, since only B implies A". Let us write this statement down : $C \neq B \implies C \nRightarrow A$. Let's interpret this equation in some other way: Put $C=A$ in this statement.Now, note that $A \neq B$ would have to be false, since if it were true, $A \nRightarrow A$ would have to be true, but this statement is clearly false! So , indeed $A=B$, which means that $A \Leftrightarrow B$, since they are identical in truth value.

So $A \Leftrightarrow B$ is equivalent to the interpretation of the statement $A $ if and only if $B$. That being said, one would wish for something more easily understandable than "iff".