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Both of those are simple consequences of the axioms. You could definitely give a proof that $-a = (-1)\cdot a$ from the axioms, but I would guess such a proof would implicitly prove those two facts.

For instance, one might do the following:

$$\begin{align*} (-1)\cdot a &= (-1)\cdot a + 0 \\ &= (-1)\cdot a + (a + -a) \\ &= ((-1)\cdot a + a) + -a \\ &= ((-1)\cdot a + 1\cdot a) + -a \\ &= (-1 + 1)\cdot a + -a \\ &= 0\cdot a + -a \\ &= (0\cdot a + 0) + -a \\ &= (0 \cdot a + (a + -a)) + -a \\ &= ((0 \cdot a + a) + -a) + -a \\ &= ((0\cdot a + 1\cdot a) + -a) + -a \\ &= ((0+1)\cdot a + -a) + -a \\ &= (1\cdot a + -a) + -a \\ &= (a + -a) + -a \\ &= 0 + -a \\ &= -a \end{align*}$$

where for illustrative purposes I've written out every... single... step.

Here's the deal though: The functional study of mathematics works on previously proven lemmas. The proof that $a+b = 0 \to b=-a$ simply follows from $$b = 0 + b = (-a + a) + b = -a + (a + b) = -a + 0 = -a,$$ which isn't bad even as I've shown every single step. Similarly, $0\cdot a = 0$ is shown by $$\begin{align*} 0\cdot a &= 0\cdot a + 0 \\ &= 0\cdot a + (a+-a) \\ &= (0\cdot a + a) + -a \\ &= (0\cdot a + 1\cdot a) + -a \\ &= (0+1)\cdot a + -a \\ &= 1\cdot a + -a \\ &= a + -a \\ &= 0\end{align*}$$ where again I've shown every single step, even though it's slightly more tedious to follow.

Once I have those two facts in my toolbox of theorems, I'm going to use them, without reproving them, because iteratively increasing the reach of your theorems is the only way we can reasonably reach anywhere useful in mathematics.

Just to finish this illustration, after I've added the above theorems to the axioms already in our toolbox, the proof is just $$a + (-1)\cdot a = 1\cdot a + (-1)\cdot a = (1+-1)\cdot a = 0\cdot a = 0$$ from which we conclude $(-1)\cdot a = -a$.

I personally prefer the latter derivation.

Both of those are simple consequences of the axioms. You could definitely give a proof that $-a = (-1)\cdot a$ from the axioms, but I would guess such a proof would implicitly prove those two facts.

For instance, one might write the following:

$$\begin{align*} (-1)\cdot a &= (-1)\cdot a + 0 \\ &= (-1)\cdot a + (a + -a) \\ &= ((-1)\cdot a + a) + -a \\ &= ((-1)\cdot a + 1\cdot a) + -a \\ &= (-1 + 1)\cdot a + -a \\ &= 0\cdot a + -a \\ &= (0\cdot a + 0) + -a \\ &= (0 \cdot a + (a + -a)) + -a \\ &= ((0 \cdot a + a) + -a) + -a \\ &= ((0\cdot a + 1\cdot a) + -a) + -a \\ &= ((0+1)\cdot a + -a) + -a \\ &= (1\cdot a + -a) + -a \\ &= (a + -a) + -a \\ &= 0 + -a \\ &= -a \end{align*}$$

where for illustrative purposes I've written out every... single... step.

Here's the deal though: The functional study of mathematics works on previously proven lemmas. The proof that $a+b = 0 \to b=-a$ simply follows from $$b = 0 + b = (-a + a) + b = -a + (a + b) = -a + 0 = -a,$$ which isn't bad even as I've shown every single step. Similarly, $0\cdot a = 0$ is shown by $$\begin{align*} 0\cdot a &= 0\cdot a + 0 \\ &= 0\cdot a + (a+-a) \\ &= (0\cdot a + a) + -a \\ &= (0\cdot a + 1\cdot a) + -a \\ &= (0+1)\cdot a + -a \\ &= 1\cdot a + -a \\ &= a + -a \\ &= 0\end{align*}$$ where again I've shown every single step, even though it's slightly more tedious to follow. (do you see where this proof appears in the middle of the first proof above?)

Once I have those two facts in my toolbox of theorems, I'm going to use them, without reproving them, because iteratively increasing the reach of your theorems is the only way we can reasonably reach anywhere useful in mathematics.

Just to finish this illustration, after I've added the above theorems to the axioms already in our toolbox, the proof is just $$a + (-1)\cdot a = 1\cdot a + (-1)\cdot a = (1+-1)\cdot a = 0\cdot a = 0$$ from which we conclude $(-1)\cdot a = -a$.

I personally prefer the latter derivation.

Both of those are simple consequences of the axioms. You could definitely give a proof that $-a = (-1)\cdot a$ from the axioms, but I would guess such a proof would implicitly prove those two facts.

For instance, one might do the following:

$$\begin{align*} (-1)\cdot a &= (-1)\cdot a + 0 + 0 \\ &= (-1)\cdot a + (a + -a) + (a + -a) \\ &= (-1)\cdot a + 1\cdot a + -a + 1\cdot a + -a \\ &= (-1 + 1 + 1)\cdot a + -a + -a \\ &= (0 + 1) \cdot a + -a + -a \\ &= 1\cdot a + -a + -a \\ &= a + -a + -a \\ &= 0 + -a \\ &= -a \end{align*}$$

where each step is more or less directly from the axioms. That is, I've combined some steps and skipped commutativity/associativity steps so it doesn't become painful, but you should be able to construct such a proof if you wanted.

Here's the deal though: The functional study of mathematics works on previously proven lemmas. The proof that $a+b = 0 \to b=-a$ simply follows from $$b = 0 + b = (-a + a) + b = -a + (a + b) = -a + 0 = -a,$$ which isn't bad even as I've shown every single step. Similarly, $0\cdot a = 0$ is shown by $$\begin{align*} 0\cdot a &= 0\cdot a + 0 \\ &= 0\cdot a + (a+-a) \\ &= (0\cdot a + a) + -a \\ &= (0\cdot a + 1\cdot a) + -a \\ &= (0+1)\cdot a + -a \\ &= 1\cdot a + -a \\ &= a + -a \\ &= 0\end{align*}$$ where again I've shown every single step, even though it's slightly more tedious to follow.

Once I have those two facts in my toolbox of theorems, I'm going to use them, without reproving them, because that's the only way we can reasonably reach anywhere useful in mathematics.

Both of those are simple consequences of the axioms. You could definitely give a proof that $-a = (-1)\cdot a$ from the axioms, but I would guess such a proof would implicitly prove those two facts.

For instance, one might do the following:

$$\begin{align*} (-1)\cdot a &= (-1)\cdot a + 0 \\ &= (-1)\cdot a + (a + -a) \\ &= ((-1)\cdot a + a) + -a \\ &= ((-1)\cdot a + 1\cdot a) + -a \\ &= (-1 + 1)\cdot a + -a \\ &= 0\cdot a + -a \\ &= (0\cdot a + 0) + -a \\ &= (0 \cdot a + (a + -a)) + -a \\ &= ((0 \cdot a + a) + -a) + -a \\ &= ((0\cdot a + 1\cdot a) + -a) + -a \\ &= ((0+1)\cdot a + -a) + -a \\ &= (1\cdot a + -a) + -a \\ &= (a + -a) + -a \\ &= 0 + -a \\ &= -a \end{align*}$$

where for illustrative purposes I've written out every... single... step.

Here's the deal though: The functional study of mathematics works on previously proven lemmas. The proof that $a+b = 0 \to b=-a$ simply follows from $$b = 0 + b = (-a + a) + b = -a + (a + b) = -a + 0 = -a,$$ which isn't bad even as I've shown every single step. Similarly, $0\cdot a = 0$ is shown by $$\begin{align*} 0\cdot a &= 0\cdot a + 0 \\ &= 0\cdot a + (a+-a) \\ &= (0\cdot a + a) + -a \\ &= (0\cdot a + 1\cdot a) + -a \\ &= (0+1)\cdot a + -a \\ &= 1\cdot a + -a \\ &= a + -a \\ &= 0\end{align*}$$ where again I've shown every single step, even though it's slightly more tedious to follow.

Once I have those two facts in my toolbox of theorems, I'm going to use them, without reproving them, because iteratively increasing the reach of your theorems is the only way we can reasonably reach anywhere useful in mathematics.

Just to finish this illustration, after I've added the above theorems to the axioms already in our toolbox, the proof is just $$a + (-1)\cdot a = 1\cdot a + (-1)\cdot a = (1+-1)\cdot a = 0\cdot a = 0$$ from which we conclude $(-1)\cdot a = -a$.

I personally prefer the latter derivation.

Both of those are simple consequences of the axioms. You could definitely give a proof that $-a = (-1)\cdot a$ from the axioms, but I would guess such a proof would implicitly prove those two facts.

For instance, one might do the following:

$$\begin{align*} (-1)\cdot a &= (-1)\cdot a + 0 + 0 \\ &= (-1)\cdot a + (a + -a) + (a + -a) \\ &= (-1)\cdot a + 1\cdot a + -a + 1\cdot a + -a \\ &= (-1 + 1 + 1)\cdot a + -a + -a \\ &= (0 + 1) \cdot a + -a + -a \\ &= 1\cdot a + -a + -a \\ &= a + -a + -a \\ &= 0 + -a \\ &= -a \end{align*}$$

where each step is more or less directly from the axioms. That is, I've combined some steps and skipped commutativity/associativity steps so it doesn't become painful, but you should be able to construct such a proof if you wanted.

Both of those are simple consequences of the axioms. You could definitely give a proof that $-a = (-1)\cdot a$ from the axioms, but I would guess such a proof would implicitly prove those two facts.

For instance, one might do the following:

$$\begin{align*} (-1)\cdot a &= (-1)\cdot a + 0 + 0 \\ &= (-1)\cdot a + (a + -a) + (a + -a) \\ &= (-1)\cdot a + 1\cdot a + -a + 1\cdot a + -a \\ &= (-1 + 1 + 1)\cdot a + -a + -a \\ &= (0 + 1) \cdot a + -a + -a \\ &= 1\cdot a + -a + -a \\ &= a + -a + -a \\ &= 0 + -a \\ &= -a \end{align*}$$

where each step is more or less directly from the axioms. That is, I've combined some steps and skipped commutativity/associativity steps so it doesn't become painful, but you should be able to construct such a proof if you wanted.

Here's the deal though: The functional study of mathematics works on previously proven lemmas. The proof that $a+b = 0 \to b=-a$ simply follows from $$b = 0 + b = (-a + a) + b = -a + (a + b) = -a + 0 = -a,$$ which isn't bad even as I've shown every single step. Similarly, $0\cdot a = 0$ is shown by $$\begin{align*} 0\cdot a &= 0\cdot a + 0 \\ &= 0\cdot a + (a+-a) \\ &= (0\cdot a + a) + -a \\ &= (0\cdot a + 1\cdot a) + -a \\ &= (0+1)\cdot a + -a \\ &= 1\cdot a + -a \\ &= a + -a \\ &= 0\end{align*}$$ where again I've shown every single step, even though it's slightly more tedious to follow.

Once I have those two facts in my toolbox of theorems, I'm going to use them, without reproving them, because that's the only way we can reasonably reach anywhere useful in mathematics.