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  "body": "\n## Proof\n\nWe compare the two functions at each point.\n\n1. Let $f \\colon A \\to B$, $g \\colon B \\to C$ and $h \\colon C \\to D$ be [[functions|0XHDTE5YMZ]].\n2. By the definition of [[composition|07HF9YCWPP]], $g \\circ f \\colon A \\to C$ and $h \\circ g \\colon B \\to D$. So $h \\circ (g \\circ f)$ and $(h \\circ g) \\circ f$ are both functions from $A$ to $D$.\n3. Let $a \\in A$. By the definition of composition, applied twice, $(h \\circ (g \\circ f))(a) = h((g \\circ f)(a)) = h(g(f(a)))$.\n4. Also $((h \\circ g) \\circ f)(a) = (h \\circ g)(f(a)) = h(g(f(a)))$.\n5. By steps 3 and 4, the two functions take the same value at each $a \\in A$.\n6. A function $u \\colon A \\to D$ is the set of pairs $(a, u(a))$ with $a \\in A$. By step 5, both functions are the same set of pairs.\n7. Two [[sets|0SJ3RX9TXA]] with the same elements are equal, by the [[axiom of extensionality|0TB16T2C97]]. Hence $h \\circ (g \\circ f) = (h \\circ g) \\circ f$.\n\n## Idea\n\nBoth ways of grouping send $a$ to $h(g(f(a)))$: first apply $f$, then $g$, then $h$. Functions that agree at every point are equal.\n",
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