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I have this proposition to prove: For all $m, n, \in\mathbb Z$: $-(m + n) = (-m) + (-n)$ Proof: \begin{align*} -(m + n) + (m + n) &= 0 + 0\\ -(m + n) + (m + n) &= (-m) + m + (-n) + n\\ (-m) + (-n) + -(m + n) + m + n &= (-m) + (-n) + m + (-m) + (-n) + n\\ -(m + n) &= (-m) + (-n) \end{align*}

What do you think? I am a bit unsure when I go from (m + n) to m + n (remove the parentheses). Thank you!

Gerry Myerson
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Johnathan
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    Yes that proof works! – Victor Feb 08 '15 at 00:55
  • An effective answer or confirmation entirely depends on what axioms you are allowed to assume (or results, etc.). – Daniel W. Farlow Feb 08 '15 at 00:57
  • You can go from ..+(m+n) to ..+m+n. Both mean same thing. Else prove $$a+(m+n)=a+m+n$$. which is there in associativity property of addition. – Harish Feb 08 '15 at 00:59
  • @Hi Harish! Yes, I worked it out: \begin{align} (m + n) &= 0 + (m + n)\ (m + n) &= (0 + m) + n\ (m + n) &= (m) + n\ (m + n) &= m + n\ \end{align} – Johnathan Feb 08 '15 at 01:02
  • @induktio I am only allowed to use the axioms for addition and multiplication (e.g. inverse, commutativity, associativity, cancellation, distributivity) :) – Johnathan Feb 08 '15 at 01:05

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