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In my course book we have something called implication arrows $\Rightarrow$ and equivalence arrows $\Leftrightarrow$ and I have never managed to understand them.

When do I know which to use and how do I know that I'm correct when I use them?

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@Theo Buehler: Thanks, updated! – Zolomon Jul 5 '11 at 10:18
I replaced the arrows by their LaTeX equivalent and changed the (induction) tag to (logic). – t.b. Jul 5 '11 at 10:20
This question deals with the "equivalence arrow". – Américo Tavares Jul 5 '11 at 11:48
up vote 3 down vote accepted

Suppose you have two propositions $P$ and $Q$.

  • $P\Rightarrow Q$ means that $P$ implies $Q$ (or if $P$, then $Q$).
  • $P\Leftrightarrow Q$ means that $P$ implies $Q$ and $Q$ implies $P$ (or $P$ if and only if $Q$).

Let me add a few simple examples.

  1. Since $\frac{2p}{2q}=\frac{p}{q}$ ($q\ne 0$) we can write $$\frac{2p}{2q}=5\Leftrightarrow \frac{p}{q}=5.\qquad (1)$$

    Its meaning is that $\frac{2p}{2q}=5$ is true if and only if $\frac{p}{q}=5$.

  2. a) If you have $y=x$, then you can say that $y^2=x^2$ and write $$y=x\Rightarrow y^2=x^2.\qquad (2)$$

    Its meaning is that if $y=x$ is true so is $y^2=x^2.$

    b) If you have $y=-x$, then you can say that $y^2=x^2$ and write

    $$y=-x\Rightarrow y^2=x^2.\qquad (3)$$

    Its meaning is that if $y=-x$ is true so is $y^2=x^2.$

    c) If you know that $y^2=x^2$ you can guarantee that $y=x$ or $y=-x$ and write

    $$y^2=x^2\Rightarrow y=x\quad \text{or}\quad y=-x.\qquad (4)$$

    Its meanig is that if $y^2=x^2$, then $y=x$ or $y=-x$.

    d) Combining $(2),(3)$ and $(4)$ you have the following equivalence

    $$y^2=x^2\Leftrightarrow y=x\quad \text{or}\quad y=-x.\qquad (5)$$

    Its meanig is that $y^2=x^2$ if and only if $y=x\quad \text{or}\quad y=-x$.

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$A \Rightarrow B$ has the meaning $B$ follows from $A,$ but this does not neccesarily holds the other way round. For instance, from $x = 13$ follows, that $x$ is prime. But you can't argue the other way round. So you can write $x = 13 \Rightarrow x\in\mathbb P,\ $ but not $x\in\mathbb P\Rightarrow x=13.$

$A\Leftrightarrow B$ has the meaning, that $A$ and $B$ are the same statement, just transformed. If $A\Leftrightarrow B$ than both $A\Rightarrow B$ and $B\Rightarrow A.$ For instance, you can say that if $x$ is even, that $x$ is not odd. This certainly also holds the other way round, so you can write $\mathrm{even}(x)\Leftrightarrow\lnot\;\mathrm{odd}(x).$

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Well, as far as I know the $\Rightarrow$ which you call the implication arrow can be used to for implying statements.

  • Example : $x^{2}-1 = 0 \Rightarrow (x+1)(x-1)=0$.

The second arrow that is $\Leftrightarrow$ I have seen it often being used for if and only if statements. That is $A \Leftrightarrow B$, means If you assume $A$ then $B$ is true and if you assume $B$ then $A$ should be true (or can be deduced.)

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There is a significant subtlety about $A \Rightarrow B$ which it is useful to clear up right at the beginning and which has been the subject of other threads. It is based on the use of implication in logic.

It is that $A \Rightarrow B$ is taken to be True except in the case that A is True and B is False.

Stated like that it can seem obvious, but this means that if A is false, then the implication is taken to be true whether B is true or false. This is a convention, which can seem counterintuitive at times, but which is very useful (which is why the convention is the way it is). Intuitively the convention encodes the fact that if A is false, the implication tells us nothing about B.

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