Given two formulas $A$ and $B$, if $A$ follows from $B$ and $B$ follows from $A$ then is it true that $A$ and $B$ are equivalent?

Question

This is true if $A$ and $B$ are statements, but formulas are statements too, so I expect the answer to be yes. But let's consider this simple example:

Formula A: $$\sum_{k=0}^{n-1}x^k=\dfrac{1-x^n}{1-x}$$

Formula B: $$\sum_{k=0}^\infty x^k=\dfrac{1}{1-x},\,\,\,\, |x|<1$$

Proof that $A\to B$: Just take the limit $n\to \infty $ in $A$ assuming $|x|<1$.

Proof that $B\to A$: $$\sum_{k=0}^\infty x^k=\sum_{k=0}^{n-1}\sum_{m=0}^\infty x^{k+mn} = \sum_{k=0}^{n-1}x^k\cdot \sum_{m=0}^\infty x^{mn}$$ From this using $B$ we get $$\dfrac{1}{1-x}=\sum_{k=0}^{n-1}x^k\cdot \dfrac{1}{1-x^n},\,\,\,\, |x|<1 $$ So we get that $A$ is true if $|x|<1$. But the condition $|x|<1$ can be dropped because this is a polynomial relation, thus $A$ follows.

But clearly $A$ and $B$ are not equivalent. Am I missing something here?

Answer 1

This is a definitional issue. The two statements that you present both imply each other (as you correctly identify). Therefore, they are equivalent to one another. They may not seem equivalent because they are saying different things, but if one of the statements is true, then it is immediately true that the other holds. Therefore, if one of these two statements doesn’t hold, then it is necessarily the case that the other formula does not hold.

In everyday English, we use equivalent to mean things that are the same. Mathematically, two statements can be saying two very different things, but they may still be equivalent to each other, as long as they are both are implied by each other (as is the case here).

Answer 2

Given two formulas $A$ and $B$, if $A$ follows from $B$ and $B$ follows from $A$ then is it true that $A$ and $B$ are equivalent?

Yes.

$$\sum_{k=0}^{n-1}x^k=\dfrac{1-x^n}{1-x}\tag A$$ $$\sum_{k=0}^\infty x^k=\dfrac{1}{1-x},\,\,\,\, |x|<1\tag B$$

Proof that $A\to B:\ldots$

Proof that $B\to A:\ldots$

But clearly $A$ and $B$ are not equivalent.

Indeed, you have proven not the equivalence of $A$ and $B,$ but rather that for $|x|<1,$ $$\sum_{k=0}^\infty x^k=\dfrac{1}{1-x}\quad\iff\quad\forall n{\in}\mathbb N\;\sum_{k=0}^{n-1}x^k=\dfrac{1-x^n}{1-x}.$$

In fact, in mathematical analysis (with its attendant axioms), since $$|x|<1\implies\sum_{k=0}^\infty x^k=\dfrac{1}{1-x}\tag{B*}$$ and $$x\ne1\;\text{ and }\;n{\in}\mathbb N\implies\sum_{k=0}^{n-1}x^k=\dfrac{1-x^n}{1-x}\tag{A*}$$ are each universally true, they are universally equivalent to each other.