The case k = 1 is trivial. Assume k = 2. The solutions of the first and second congruences are

x = a₁ + l₁ m₁,   with   l₁ integral,

and

x = a₂ + l₂m₂,   with   l₂ integral.

For a solution of both congruences we therefore have

a₁ + l₁m₁ = a₂ + l₂m₂,

i.e.,

l₁m₁ - l₂m₂ = a₂ - a₁.

Since gcd(m₁,m₂) = 1 there exist integers p₁, p₂ such that

p₁m₁ + p₂m₂ = 1.

Therefore, a solution of both congruences is

x = a₁ + (a₂ - a₁)p₁m₁ = a₂ - (a₂ - a₁)p₂m₂.

By the induction hypothesis the system

x = a_i mod m_i,   i = 1, 2, ..., k - 1,

has a solution, a say. Now consider the system of two congruences:

x = a mod m₁m₂ ··· m_{k - 1},

x = a_k mod m_k.

The numbers m₁m₂ ··· m_{k - 1} and m_k are relatively prime, so the case k = 2 tells us that this system has a solution. This solution is also a solution of the original system.

If x and y are two solutions, then x - y is divisible by m₁, m₂, ... , m_k and therefore by the product m₁ m₂ ··· m_k (since all m_i are pairwise relatively prime). In other words, the solutions x and y are equal modulo m₁ m₂ ··· m_k.