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I was wondering why do some people use redundant axioms in definitions?

If you just expand $(a+1)(b+1)=(a+1)b+a+1=ab+b+a+1$ $(a+1)(b+1)=a(b+1)+b+1=ab+a+b+1$. Hence, $ab+a+b+1=ab+b+a+1$, then cancel ab and 1. Then, you get it's commutative for free.

Why do we then assume it?

Also, is it even stronger. If we drop the condition that 1 is in R, then can we still deduce a+b=b+a? As clearly the argument just done falls apart if you haven't got 1 in R.

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    @simplicity: To reinforce Qiaochu's point, I'll mention that there is an axiomatisation of the theory of groups that only has one operation and one axiom: see [here](http://www.cs.unm.edu/~mccune/projects/gtsax/). It's completely useless.2011-12-27

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Without a $1$, you cannot deduce the condition. Take $(R,\cdot,\times)$ where $(R,\cdot)$ is any nonabelian group with identity element $e$, and let $a\times b = e$ for all $a$ and $b$ (the "zero multiplication ring"). This satisfies all the axioms of a ring without $1$, except for commutativity of the first operation. So $a+b=b+a$ cannot be deduced from the rest of the axioms of a ring without a $1$.

Another method to deduce commutativity when there is a $1$ is to consider $(a+b)(1+1)$ and distribute both ways: $\begin{align*} (a+b)(1+1) &= (a+b)+(a+b) = a+b+a+b,\\ (a+b)(1+1) &= a(1+1) + b(1+1) = a+a+b+b. \end{align*}$ From $a+b+a+b = a+a+b+b$ we deduce $b+a=a+b$.

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Without $1$, commutativity of $+$ does not come for free. Suppose $R$ is a non-abelian group under $+$ and define $x \cdot y=0$ for all $x$, $y$. The ring axioms, other than commutativity of $+$, are satisfied.