[proofplan] We prove the identity directly from the divisibility definition of [greatest common divisor](/page/Greatest%20Common%20Divisor). Set $g=\gcd(a,b)$ and $h=\gcd(g,c)$. First we show that $h$ divides each of $a,b,c$, using $h \mid g$ and $g \mid a,b$. Then we show that every common divisor of $a,b,c$ divides $h$, by passing through the defining maximal divisibility property of $g$ and then of $h$. [/proofplan]

[step:Define the two binary gcds and verify they are available] Let \begin{align*} g = \gcd(a,b). \end{align*} Since $(a,b) \ne (0,0)$ and greatest common divisors are nonnegative, we have $g \in \mathbb{Z}_{>0}$. In particular $(g,c) \ne (0,0)$, so the binary greatest common divisor \begin{align*} h = \gcd(g,c) \end{align*} is defined and satisfies $h \in \mathbb{Z}_{\ge 0}$. [/step]

[step:Show that $h$ has the defining divisibility property of $\gcd(a,b,c)$]Since $h=\gcd(g,c)$, the definition of the binary gcd gives $h \mid g$ and $h \mid c$. Since $g=\gcd(a,b)$, the definition of the binary gcd gives $g \mid a$ and $g \mid b$. Divisibility is transitive in $\mathbb{Z}$, so $h \mid a$ and $h \mid b$. Therefore $h$ divides each of $a,b,c$. Now let $r \in \mathbb{Z}$ be any common divisor of $a,b,c$. Then $r \mid a$ and $r \mid b$, so the defining property of $g=\gcd(a,b)$ gives $r \mid g$. Since also $r \mid c$, the integer $r$ is a common divisor of $g$ and $c$. The defining property of $h=\gcd(g,c)$ therefore gives $r \mid h$.[/step]

[guided]We must prove that $h$ satisfies exactly the definition assigned to $\gcd(a,b,c)$ in the theorem statement. There are two requirements: first, $h$ must divide all three integers $a,b,c$; second, every common divisor of $a,b,c$ must divide $h$. For the first requirement, use the two binary gcd definitions in sequence. Since $h=\gcd(g,c)$, we have \begin{align*} h \mid g \end{align*} and \begin{align*} h \mid c. \end{align*} Since $g=\gcd(a,b)$, we have \begin{align*} g \mid a \end{align*} and \begin{align*} g \mid b. \end{align*} The relation $h \mid g$ and $g \mid a$ implies $h \mid a$ by transitivity of divisibility in $\mathbb{Z}$; similarly, $h \mid g$ and $g \mid b$ imply $h \mid b$. Together with $h \mid c$, this proves that $h$ is a common divisor of $a,b,c$. For the second requirement, take an arbitrary integer $r \in \mathbb{Z}$ such that $r \mid a$, $r \mid b$, and $r \mid c$. Because $r$ divides both $a$ and $b$, the defining greatest-divisor property of $g=\gcd(a,b)$ gives \begin{align*} r \mid g. \end{align*} Together with $r \mid c$, this says that $r$ is a common divisor of $g$ and $c$. Applying the defining greatest-divisor property of $h=\gcd(g,c)$ gives \begin{align*} r \mid h. \end{align*} Thus every common divisor of $a,b,c$ divides $h$, which is the key maximality condition for the three-term gcd.[/guided]

[step:Conclude by uniqueness of the three-term greatest common divisor] The previous step proves that $h$ is a nonnegative integer dividing $a,b,c$ and that every common divisor of $a,b,c$ divides $h$. Since $(a,b) \ne (0,0)$, the three integers $a,b,c$ are not all zero, so the defining uniqueness of $\gcd(a,b,c)$ applies. Hence \begin{align*} h=\gcd(a,b,c). \end{align*} Recalling that $h=\gcd(g,c)$ and $g=\gcd(a,b)$, we obtain \begin{align*} \gcd(a,b,c)=\gcd(\gcd(a,b),c). \end{align*} This is the desired identity. [/step]