Show that if a≡b(mod m) and c≡d(mod m), where a, b, c, d ϵ Z with m≥2, then a-c≡b-d(mod m).

Given an integer $m>1$, called a modulus, two integers are said to be congruent modulo $m$ , if $m$ is a divisor of their difference (i.e., if there is an integer $k$ such that $a-b=km$). Congruence modulo $m$ is denoted: $a\equiv b(\mod m)$.

If $a\equiv b(\mod m)$ and $c\equiv d(\mod m)$, then there exist integers $k$ and $s$ such that $a-b=km$ and $c-d=sm$. Then $(a-b)-(c-d)=km-sm$. It follows that $(a-c)-(b-d)=(k-s)m$ where $k-s$ is an integer as a difference of two integers. Therefore,

$a-c\equiv b-d(\mod m)$.