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  "body": "\n## Proof\n\nWe argue by [[induction|0ASJVEFA9J]] on $c$.\nLet $S$ be the [[successor function|0832XVR512]].\nThe [[addition|0W5DSDFQ4G]] satisfies $x + 0 = x$ and $x + S(y) = S(x + y)$ for all natural numbers $x$ and $y$.\nFix natural numbers $a$ and $b$.\n\n1. For $c = 0$: the definition gives $(a + b) + 0 = a + b$.\n2. Also $b + 0 = b$, so $a + (b + 0) = a + b$. Thus $(a + b) + 0 = a + (b + 0)$.\n3. Suppose that $(a + b) + c = a + (b + c)$ for some natural number $c$.\n4. By the definition, $(a + b) + S(c) = S((a + b) + c)$.\n5. By the hypothesis of step 3, $S((a + b) + c) = S(a + (b + c))$.\n6. By the definition, $S(a + (b + c)) = a + S(b + c)$.\n7. By the definition again, $S(b + c) = b + S(c)$, so $a + S(b + c) = a + (b + S(c))$.\n8. Steps 4 to 7 give $(a + b) + S(c) = a + (b + S(c))$.\n9. Since $S(c) = c + 1$, induction gives $(a + b) + c = a + (b + c)$ for every natural number $c$. As $a$ and $b$ were arbitrary, the law holds for all $a$, $b$ and $c$.\n\n## Idea\n\nThe recursive definition of addition peels one successor off the last number at a time.\nOn both sides, this successor moves to the outside in the same way.\nSo the case $S(c)$ reduces to the case $c$.\n",
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        "locator": "Proof of associativity",
        "url": "https://en.wikipedia.org/wiki/Proofs_involving_the_addition_of_natural_numbers"
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    "tag": "KBVJPBJXFD",
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