Subtraction of Subring is Subtraction of Ring

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Theorem

Let $\struct {R, +, \circ}$ be an ring.

For each $x, y \in R$ let $x - y$ denote the subtraction of $x$ and $y$ in $R$.


Let $\struct {S, + {\restriction_S}, \circ {\restriction_S}}$ be a subring of $R$.

For each $x, y \in S$ let $x \sim y$ denote the subtraction of $x$ and $y$ in $S$.


Then:

$\forall x, y \in S: x \sim y = x - y$


Proof

Let $x, y \in S$.

Let $-x$ denote the ring negative of $x$ in $R$.

Let $\mathbin \sim x$ denote the ring negative of $x$ in $S$.

Then:

\(\displaystyle x \sim y\) \(=\) \(\displaystyle x \mathbin {+ {\restriction_S} } \paren {\mathbin \sim y}\) Definition of Ring Subtraction
\(\displaystyle \) \(=\) \(\displaystyle x + \paren {\mathbin \sim y}\) Definition of Addition on Subring
\(\displaystyle \) \(=\) \(\displaystyle x + \paren {-y}\) Negative of Subring is Negative of Ring
\(\displaystyle \) \(=\) \(\displaystyle x - y\) Definition of Ring Subtraction

$\blacksquare$