I recently had a lecture in a proof class where we were proving a Theorem of the form
Theorem 6.18 Let condition be true. Then the following are equivalent.
a) Statement 1
b) Statement 2
c) Statement 3
d) Statement 4
e) Statement 5
When we began to prove the theorem in the following manner of showing $a\implies b$ and then $b \implies c$ and so forth until $e\implies a$
Is this valid? It seems circular in essence since we assumed $a$ to prove everything.
On
Generally, these statements tend be predicates. That is, they are statements that can be true or false, given some input. For instance, given a square nxn matrix A, the following are equivalent:
S1. A is invertible
S2. det(a) is not 0
S3. dim(nullspace(A)) = 0
S4. dim(columnspace(A)) = n
S5. In row-reduced form, A has n pivot columns
You can think of S1 ... S5 as being functions that return some truth value. Put in different matrices A, and you'll get different output. S1(I) = true; the identity is indeed invertible. Saying they're equivalent means that while the output might vary for different A, it doesn't vary for different functions; S1(A) = S2(A) = S3(A) = S4(A) = S5(A).
So when proving a theorem of this sort, S1 ... S5 are not being proved to be true; they clearly are not always true (there are matrices that are not invertible). What is being proven is that S1 .. S5 are equal to each other. Proving that two predicates are equal can be done by proving that the first implies the second, and the second implies the first. For five different statements, that's a total of 20 different implications. However, we can take a shortcut of only showing five implications; if we have a cycle of each statement implying the next, then a chain of implications exist from any one to another.
Another way of thinking about it is that if one statement implies another, then it is at least as strong as the statement it implies. Just as showing
$a \geq b \geq c \geq d \geq e \geq a$ shows that a = b = c = d = e
showing that S1 is at least as strong as S2, which is at least as strong as S3, which is at least as strong as S4, which is at least as strong as S5, which is at least as strong as S1, shows that S1 = S2 = S3 = S4 = S5.
On
" It seems circular in essence since we assumed $a$ to prove everything."
Except you didn't actually prove anything is actually true (all those statements could be utterly false for all we know). You proved $a \implies$ everything else is true.
If you want to prove a statement: $a \implies K$. It is perfectly fine to "assume" $a$ is true. Because you aren't actually assuming $a$ will always be true. You are simply proving that in the cases when $a$ happens to be true that $K$ will follow. That is not assuming that $a$ is true. $a$ doesn't need to be true for $a \implies K$ to be true.
A statement: The following are equivalent; $a,b,c,d,e$ is a statement that means (among other things):
if $a$ happens to be true then so will $b,c,d,e$. (It also means if any of the other four happen to be true than all five will be.)
"The following are equivalent" means that the following statements share the same truth value. That is, if any one of the statements is true then all are true, and if any one is false then all are false. So you can prove $a\implies b$ and $b\implies c$ etc. to obtain a loop of implications, which allows you to go from any one statement to any of the others. For instance, if $b$ is true, then in the proof we showed $b\implies c\implies d\implies e\implies a$ so that the other four statements are true.
The statement of the theorem is not saying "the following are all true", and so the proof is not showing this. There is a difference between a statement being true and it being equivalent to another statement. For example, consider the well-known result:
This statement gives a logical equivalence between $(1)$ and $(2)$, however it is not claiming that $(1)$ and $(2)$ always hold; indeed not every function is continuous, but every continuous function satisfies $(2)$ and vice-versa.