Proof of $a - b = - (b - a)$, Discrete Math

Question

Prove that a - b = - (b - a) with the following axioms:

“Associativity of +”: (a + b) + c = a + (b + c)

“Associativity of ·”: (a · b) · c = a · (b · c)

“Symmetry of +”: a + b = b + a

“Symmetry of ·”: a · b = b · a

“Additive identity” “Identity of +”: 0 + a = a

“Multiplicative identity” “Identity of ·”: 1 · a = a

“Distributivity of · over +”: a · (b + c) = a · b + a · c

“Zero of ·”: a · 0 = 0

“Unary minus”: a + (- a) = 0

“Subtraction”: a - b = a + (- b)

I have no idea how to distribute the - sign at - (b - a) in to the parentheses.

Any help would be appreciated. Thank you so much in advance.

Note: You can't add the same number/variable to the both sides or can't say a - b + b - a = 0 by unary minus.

Clarification for note:

We use an online system to enter our proofs and we start with either one side or both sides. Exm: Proof: a - b then we do the next step under it Exm: a + (- b) - Subtraction however we cannot start by one side and make up another equality

Exm: a - b + b -a = 0 - by unary minus is not allowed we cannot put equality sign if we started with only one side.

When we start with one side the goal is to get the same exact thing on the other side. Exm: in this case after the proofs steps I need to end up with - (b - a)

Answer 1

$a-b\overset{Additive \ Identity}{=}$

$0+(a-b)\overset{Unary \ Minus}{=}$

$((b-a) + (-(b-a)))+(a-b)\overset{Symmetry \ of \ +}{=}$

$((-(b-a))+(b-a))+(a-b)\overset{Associativity \ of \ +}{=}$

$(-(b-a))+((b-a)+(a-b))\overset{Subtraction \ x \ 2}{=}$

$(-(b-a))+((b+(-a))+(a+(-b)))\overset{Associativity \ of \ +}{=}$

$(-(b-a))+(b+((-a)+(a+(-b))))\overset{Associativity \ of \ +}{=}$

$=(-(b-a))+(b+(((-a)+a)+(-b)))\overset{Symmetry \ of \ +}{=}$

$=(-(b-a))+(b+((a+(-a))+(-b)))\overset{Unary \ Minus}{=}$

$=(-(b-a))+(b+(0+(-b)))\overset{Additive \ Identity}{=}$

$=(-(b-a))+(b+(-b))\overset{Unary \ Minus}{=}$

$=(-(b-a))+0\overset{Symmetry \ of \ +}{=}$

$=0+(-(b-a))\overset{Additive \ Identity}{=}$

$=(-(b-a))$

Answer 2

Use distributivity of multiplication over addition: We have that $a\cdot(b+c)=a\cdot\,b+a\cdot\,c$ therefore $-(b-a)=-1(b-a)=-1\cdot\,b+-1\cdot\,-a=-b+a=(a-b).$

Answer 3

Since $x=-y$ is equivalent to $x+y=0$, just evaluate $a-b+b-a$ with repeated use of addition's commutativity and associativity, and $-z+z=z-z=0$. The sum is $a+0-a=a-a=0$, as required.