Ring Subtraction equals Zero iff Elements are Equal

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Theorem

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$

Then:

$\forall a, b \in R: a - b = 0_R \iff a = b$

where $a - b$ denotes ring subtraction.


Proof

\(\displaystyle a - b\) \(=\) \(\displaystyle 0_R\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle a + \paren {-b}\) \(=\) \(\displaystyle 0_R\) Definition of Ring Subtraction
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \paren {a + \paren {-b} } + b\) \(=\) \(\displaystyle 0_R + b\) Cancellation Laws
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle a + \paren {b^{-1} + b}\) \(=\) \(\displaystyle 0_R \circ b\) Group Axiom $G \, 1$: Associativity
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle a\) \(=\) \(\displaystyle b\) Group Axiom $G \, 2$: Identity and Group Axiom $G \, 3$: Inverses

$\blacksquare$


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