# Teaching "if...then" and "the following are equivalent"

What do you say to the following way of teaching "if...then" and "the following are equivalent"? Has somebody ever taught it like this?

An implication $$(A {\implies} B)$$ can be viewed as asserting that $$B$$ is at least as true as $$A.$$ Thus, if $$(A {\implies} B)$$ and $$(B{\implies} A),$$ then $$A$$ and $$B$$ have the same truth value ("at least as true" in both directions).

Also, under this interpretation, it is easy to see that it suffices to prove the cyclic chain of implications $$(A_1 {\implies} A_2 {\implies} \dots {\implies} A_n {\implies} A_1)$$ in order to show that $$A_1, \ldots, A_n$$ are equivalent, that is, have the same truth value.

• Present this as a generalization of $A\Leftrightarrow B$. Your chain is the same as $A_1\Leftrightarrow A_2,\ A_2\Leftrightarrow A_3,\cdots, A_n\Leftrightarrow A_1$ but organized in a way tailored for proving certain "big concepts." Sep 28, 2016 at 19:54
• Mathematicians routinely use the terms stronger statement and weaker statement. “At least as true” is a bit confusing: $A \Rightarrow B$ seems at least as true as $B \Rightarrow A$, unless you mean one of them implies the other one, because each has three Ts in its truth table. Well that’s not what you mean of course, so now I have to struggle to remember that there’s some technical restriction in comparing the amounts of truth in two propositions. However, I do teach stronger and weaker because it’s common jargon. Jul 25, 2022 at 13:10
• "as least as true as": by formulating it like this, you are using the equivalence between $A \implies B$ and $A \le B$. Although this is mathematically correct, this is very hazardous for students, starting in that matter. Dec 21, 2023 at 8:24
• What's wrong with the standard $A \implies B ~\equiv ~ \neg(A \land \neg B)$ ? Dec 22, 2023 at 15:53

Maybe Terence Tao's logic section of his text "Analysis 1" is an example. He writes:

One can also think of the statement “if X, then Y ” as “Y is at least as true as X” - if X is true, then Y also has to be true, but if X is false, Y could be as false as X, but it could also be true. This should be compared with “X if and only if Y ”, which asserts that X and Y are equally true.

So there he explains why $(A\implies B)\quad\land\quad (B\implies A)$ is the same as saying that $A$ and $B$ have the same truth value. He also gives the following exercise:

Suppose you know that X is true if and only if Y is true, and you know that Y is true if and only if Z is true. Is this enough to show that X, Y, Z are all logically equivalent? Explain.

Using the information given in the text (that "If A, then B" means "B is at least as true as A"), the solution you have given is obvious.

I don't know that that's the most common way of teaching it, however, the idea that "if A then B" represents an ordering of truth values such that B is at least as true as A is mathematically and conceptually sound.

It's an interesting idea. In programmers' terms, if you count 0 as your "truth" value and 1 as your "false" value (which is how it works in some programming languages), then:

$A => B$

means "A is equal to or greater than B."

A >= B


;)

So if A is 0, then B has to be 0 as well (true) because it can't be anything smaller.

Likewise if B is 1 (false), then A can't be anything greater, so must be equal (also false).

But if A is 1, B could satisfy the comparison by being either 0 or 1.

It's a very interesting (and humorous!) thought experiment.

I would NOT recommend it as your first pedagogical entry to the subject of logical implication.

Bring it up after your students are comfortable with the idea of logical implication, and bring it up labeled clearly as an unusual way of looking at it, not generally agreed upon. (My advice.)

What do you say to the following way of teaching "if...then"?

An implication $$(A {\implies} B)$$ can be viewed as asserting that $$B$$ is at least as true as $$A.$$

Instead of suggesting that the truth of $$A$$ lies on a spectrum $$(A$$ after all is unambiguously either true or false), why not explicitly write $$\text{for each }x,\;\Big(A(x){\implies}B(x)\Big),$$ or just $$A(x){\implies}B(x),$$ and say that $$\text“A(x)$$ being true forces $$B(x)$$ to be true” or, more clearly, “whenever $$A(x)$$ is true, $$B(x)$$ is necessarily true”, that is, every value of $$x$$ that satisfies $$A(x)$$ also satisfies $$B(x).$$

An implication $$(A {\implies} B)$$ can be viewed as asserting that $$B$$ is at least as true as $$A.$$

If set $$A$$ contains every value of $$x$$ that satisfies $$A(x),$$ and likewise for $$B(x),$$ then set $$B$$ contains at least the elements of set $$A$$ (i.e., $$A\subseteq B$$).

Thus, if $$(A {\implies} B)$$ and $$(B{\implies} A),$$ then "at least as true" in both directions

In this case, $$A(x)$$ and $$B(x)$$ have the same solution set.

What do you say to the following way of teaching "the following are equivalent"?

it suffices to prove that $$(A_1 {\implies} A_2 {\implies} \dots {\implies} A_n {\implies} A_1)$$

This is intuitive and symmetrical. Equivalently, it suffices to show that

• $$A_1\implies A_2\land A_3\land A_4\land\ldots\land A_n$$
• $$\lnot A_1\implies\lnot A_2\land \lnot A_3\land \lnot A_4\land \ldots\land\lnot A_n$$

(assume, without loss of generality, that $$A_3$$ is true; then, by contraposition, so is $$A_1;$$ consequently, so are the remaining statements).