Common Divisor Divides Integer Combination/General Result

Theorem
Let $c$ be a common divisor of a set of integers $A := \set {a_1, a_2, \dotsc, a_n}$.

That is:
 * $\forall x \in A: c \divides x$

Then $c$ divides any integer combination of elements of $A$:


 * $\forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
 * $\forall x \in \set {a_1, a_2, \dotsc, a_n}: c \divides x \implies \forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$

Basis for the Induction
$\map P 2$ is the case:
 * $\forall x \in \set {a_1, a_2}: c \divides x \implies \forall x_1, x_2 \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2}$

This is demonstrated to be true in Common Divisor Divides Integer Combination.

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\forall x \in \set {a_1, a_2, \dotsc, a_k}: c \divides x \implies \forall x_1, x_2, \dotsc, x_k \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k}$

from which it is to be shown that:
 * $\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x \implies \forall x_1, x_2, \dotsc, x_k, x_{k + 1} \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }$

Induction Step
This is the induction step:

Let :
 * $\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x$

We have that:
 * $c \divides a_{k + 1} \implies \forall x_{k + 1} \in \Z: c \divides a_{k + 1} x_{k + 1}$

and we have that:
 * $\forall x \in \set {a_1, a_2, \dotsc, a_k}: c \divides x \implies \forall x_1, x_2, \dotsc, x_k \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k}$

Let $x_1, x_2, \dotsc, x_k \in \Z$ be arbitrary.

Let $d = a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k$.

Then:

But $x_1, x_2, \dotsc, x_k \in \Z$ are arbitrary, and so:


 * $\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x \implies \forall x_1, x_2, \dotsc, x_k, x_{k + 1} \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 2}: \forall x \in \set {a_1, a_2, \dotsc, a_n}: c \divides x \implies \forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$