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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

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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.)

Answer 4

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 $A.$

This alludes to the meaning of $(A {\implies} B)$ as $\text“A$ being true forces $B$ to be true”.

But statement $A$ is exclusively either true or it's false; so, what's this “at least as true as” degree-of-truth idea all about??

It is less confusing down the road—and anyway more accurate—to present $$\text{for each }x,\;\Big(A(x){\implies}B(x)\Big),$$ explain that $A(x)$ and $B(x)$ is each a conditional statement (e.g., $x{=}{-}3$ and $x^2{=}9,$ respectively), whose truth depends on its input $x,$ and only here assert, $\text“B(x)$ is at least as true as $A(x)\text”.$

For simplicity, it is okay to just write $$A(x){\implies}B(x),$$ with the “for each $x$” implicitly understood. This (implicitly universally-quantified) implication means that every value of $x$ that satisfies $A(x)$ also satisfies $B(x).$ In other words, if sets $A$ and set $B$ respectively contains every value of $x$ that satisfies $A(x)$ and $B(x),$ then set $B$ contains at least the elements of set $A$ (i.e., $A\subseteq B$); it is in this sense that $B(x)$ is at least as true as $A(x).$

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

Yes; 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 $(A_1 {\implies} A_2 {\implies} \dots {\implies} A_n {\implies} A_1)$

Yes, this is good since it's intuitive and symmetrical.

Just for interest, a logically equivalent alternative is to show that

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

and that

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

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