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I was reading from the book Apostol calculus, There is this theorem

Possibility of Subtraction : Given $a$ and $b$, there is exactly one $x$ such that $a+x=b$.

The proof is - Given $a$ and $b$, choose $y$ so that $a + y = 0$ and let $x = y + b$. Then $a+x = a+(y+b) = (a+y) +b = 0+b = b$. Therefore there is at least one $x$ such that $a + x = b$. I understand till here but then its written by Theorem 1.1 there is at-most one such $x$. Hence there is exactly one $x$.

Theorem 1.1 is If $ a + b = a + c $ then b = c.

How does theorem 1.1 tells us there is at-most one x

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Suppose $x$ and $y$ are such that $a+x = b$ and $a+y=b$. Then $a+x = a+y$, so by the quoted Theorem 1.1, we get $x=y$.

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