Broadly speaking, we generally consider two statements to be equivalent, whenever one implies the other and vice versa. Wolfram's page on equivalent touches on this definition.We'll write it as follows:

\[\begin{equation*}

P \iff Q \text{ provided that }P \implies Q \land Q \implies P

\end{equation*}\]

So let's begin by constructing a formal system. We'll begin as we did before and assume the following two definitions. We'll call this system $S$.

\[\begin{align*}

\vec{p} &= m \vec{v} \tag{I: Definition of Momentum} \\

\vec{a} &= \frac{d\vec{v}}{dt} \tag{II: Definition of Acceleration}

\end{align*}\]

Previously we observed that we were not able to prove that $\vec{F}=m\vec{a}$ follows from Newton's 2nd Law with just these assumptions. As a result, $\vec{F}=m\vec{a}$ is

*not equivalent*to Newton's 2nd Law in system $S$.

So let's augment system $S$ with on additional assumption. Let's consider system $S^*$ which is as follows:

\[\begin{align*}

\vec{p} &= m \vec{v} \tag{I: Definition of Momentum} \\

\vec{a} &= \frac{d\vec{v}}{dt} \tag{II: Definition of Acceleration} \\

\frac{dm}{dt} &= 0 \tag{III: Invariance of Mass}

\end{align*}\]

We had previously proved that if Newton's Law holds good then $\vec{F}=m\vec{a}$ follows from these assumptions. So in order to prove equivalence, we will need to show that given $\vec{F}=m\vec{a}$, Newton's 2nd Law follows. This is simple enough to show.

\[\begin{align*}

\vec{F} &= m \vec{a} \tag{1: Assumption}\\

&= m \frac{d \vec{v}}{dt} \tag{2: from 1 and II} \\

\frac{d\vec{p}}{dt} &= m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt} \tag{3: from I} \\

m\frac{d\vec{v}}{dt} &= \frac{d\vec{p}}{dt} - \vec{v}\frac{dm}{dt}\tag{4: from 3} \\

\vec{F} &= \frac{d\vec{p}}{dt} - \vec{v}\frac{dm}{dt}\tag{5: from 2 and 4} \\

&= \frac{d\vec{p}}{dt} \tag{6: from III and 5} \end{align*}\]

So we have thus shown that Newton's 2nd Law is equivalent to $\vec{F}=m\vec{a}$

__$S^*$.__

*in system*It can also be shown that Newton's 2nd Law is equivalent to $\vec{F}=m\vec{a}$ in system $S^+$ which we define as the following three propositions:

\[\begin{align*}

\vec{p} &= m \vec{v} \tag{I: Definition of Momentum} \\

\vec{a} &= \frac{d\vec{v}}{dt} \tag{II: Definition of Acceleration} \\

\vec{F} = 0 &\iff \frac{d\vec{v}}{dt} = 0 \tag{III: Newton's First Law}

\end{align*}\]

As a result we can ask the following question: Is Newton's 2nd Law equivalent to $\vec{F}=m\vec{a}$? The answer to that is "yes" in systems $S^*$ and $S^+$ but not in $S$.

A similar thing happens when consider system $T$:

\[\begin{align*}

\vec{p} &= m \vec{v} \tag{I: Definition of Momentum} \\

\vec{a} &= \frac{d\vec{v}}{dt} \tag{II: Definition of Acceleration} \\

\vec{F} &= \frac{d\vec{p}}{dt} \tag{III: Newton's 2nd Law}

\end{align*}\]

In system $T$, it can be shown that Newton's First Law is equivalent to the Invariance of Mass. I'll leave out the details of the proof.

So what does it mean to say that two statements are equivalent? I would like to modify the initial proposed definition.

$$\text{Statements }P\text{ and }Q\text{ are equivalent in system }S\text{ provided that } P \implies Q \text{ in }S \land Q \implies P \text{ in }S$$

The conclusion is that [logical] equivalence is

*system dependent*.

Now there's perhaps a related point here as to whether or not we are talking about logical equivalence or some other form of equivalence. I believe the results will end up being the same. The only difference is how we regard the status of the system in question (whether its axioms and rules are "purely logically" or not).

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