Proof of $a - (b + c) = a - b - c$, Discrete Mathematics

Question

Prove that a - (b + c) = a - b - c 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)

So far what I did:

Proof:

a - (b + c)

a - (b + c) + 0 by Identity of +

a - (b + c) + (- (b + c) + (b + c)) by Unary minus

However, I can't get rid of the - sign in front of the - (b + c) what should I do?

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

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 - c then we do the next step under it Exm: a + (- b) + (- c) 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 a - (b + c)

Answer 1

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

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

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

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

$(((a-b)-c)+(b+c))+(-(b+c))\overset{Subtraction \ x \ 3}{=}$

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

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

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

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

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

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

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

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

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

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

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

$a-(b+c)$

Answer 2

Hint: Start with $((a-b)-c) +(b+c)$.

Answer 3

I'd prove that $x+y = 0\implies y=-x$ first.

Then I'd show that $(b+c) + (-b) + (-c) = 0$.

That'd mean that $-(b+c) = (-b) + (-c)$.

Then $a +(-(b+c)) = a+(-b) +(-c)$.

I trust you can fill in the details.

Answer 4

Proof:

a - b - c start by right side

a - b - c + 0 by Identity of +

a - b - c + (- (b + c) + (b + c)) by Unary minus

a + - b + - c + - (b + c) + (b + c) by Subtraction

a + - b + - c + (b + c) + - (b + c) by Symmetry of +

(a + ((- b + b) + (- c + c))) + - (b + c) by Associativity of +

a + 0 + 0 + - (b + c) by Unary minus

a + - (b + c) by Identity of +

a - (b + c) by Subtraction