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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?

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  • $\begingroup$ Perhaps an example would be helpful? For $n\times n$ matrix $A$, $\det(A)\neq 0$ if and only if $A$ is invertible if and only if the columns of $A$ are linearly independent if and only if $\ker(A)=\{0\}$ if and only if... If you were to take any one of those properties listed as a hypothesis, you can prove all of the rest must be true as well. $\endgroup$ – JMoravitz Mar 11 '16 at 1:24
  • $\begingroup$ Look up "truth tables" and this will all start to make sense. $\endgroup$ – goblin Mar 11 '16 at 1:39
  • $\begingroup$ It simply means that $A$ and $B$ either both hold or else both fail. It doesn't imply any sort of deeper connection between $A$ and $B$, and there's no other meaning behind it; it's just a statement about a truth table. $\endgroup$ – anomaly Mar 13 '16 at 7:53
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    $\begingroup$ What do you mean by "makes" $A$ true? "The statement A if and only if B sounds like A is true precisely in case B is true, i.e., if B then A, and if ¬B then ¬A (or equivalently, if A then B)." This is correct. $\endgroup$ – littleO Mar 13 '16 at 8:04
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    $\begingroup$ @littleO Well, what I was getting at is that A iff B sounds like one and only one thing (i.e., B and nothing else) is sufficient for A. $\endgroup$ – user185744 Mar 13 '16 at 9:51
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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.

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  • $\begingroup$ Do you mean case (2)? Also, I think I may be interpreting "iff" incorrectly since my interpretation would require one to write "and conversely" to include the other implication. A if and only if B sounds like A happens/is true whenever but only whenever B is true, i.e., like a stronger form of if. $\endgroup$ – user185744 Mar 11 '16 at 2:13
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    $\begingroup$ @K.Hotz, that's exactly right. $\endgroup$ – goblin Mar 11 '16 at 2:35
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    $\begingroup$ @K.Hotz, I prefer iff, because its more consistent. Compare: let $f : \mathbb{R} \rightarrow \mathbb{R}$ be given as follows: $$f(x) = x^2.$$ Let $P : \mathbb{R} \rightarrow \mathbb{B}$ be given as follows: $$P(x) \iff x^2 \leq 2.$$ $\endgroup$ – goblin Mar 11 '16 at 2:51
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    $\begingroup$ @K.Hotz, the way I see it, $\iff$ is a hell of a lot like $=$, and $\implies$ is a hell of a lot like $\leq$. My argument is that since you wouldn't define $f$ by writing $f(x) \geq x^2$, so why do this for definitions? $\endgroup$ – goblin Mar 11 '16 at 2:53
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    $\begingroup$ I had a nightmare yesterday because of your profile picture. BTW thanks a lot for such a nice answer (+1). $\endgroup$ – user231343 Mar 19 '16 at 22:18
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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 $A \iff B$

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  • $\begingroup$ Right, but does "iff" mean that A precisely in case B? $\endgroup$ – user185744 Mar 11 '16 at 1:43
  • $\begingroup$ @K.Hotz, yes. $\;\!$ $\endgroup$ – goblin Mar 11 '16 at 2:03
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"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 $A\implies B$.

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  • $\begingroup$ Right, but "A if and only if B" sounds as though we mean that A is true precisely in case B is true. Is this correct, and if so how does this interpretation relate to two things implying one another? $\endgroup$ – user185744 Mar 11 '16 at 1:55
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    $\begingroup$ It is right if we take "precisely" to mean "only"; in other words, if A is true when B is true and is false when B is false. Then we have two statements as in the answer, and we get mutual implication. $\endgroup$ – user321773 Mar 11 '16 at 2:25
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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".

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  • $\begingroup$ Yes, how I am interpreting the "only if" part is exactly as you write. Except it sounds as though we should write "and conversely" at the end. $\endgroup$ – user185744 Mar 11 '16 at 2:04
  • $\begingroup$ Right. That's good. $\endgroup$ – астон вілла олоф мэллбэрг Mar 11 '16 at 2:36

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