By the definition of Congruence, there are integers s and t such that,

a-b = sn...(1)

and

c-d=tn...(2)

(1)-(2)

(a-b)-(c-d)=sn-tn

(a-b)-(c-d)=n(s-t)

Both s and t are integers so s−t is also integer

put s−t=u

(a-b)-(c-d)=nu

By defination of Congruence

\(\displaystyle{\left({a}-{b}\right)}\equiv{\left({c}-{d}\right)}{\left(\text{mod}{n}\right)}\)

Hence Proved,

\(\displaystyle{\left({a}-{c}\right)}\equiv{\left({c}-{d}\right)}{\left(\text{mod}{n}\right)}\)

a-b = sn...(1)

and

c-d=tn...(2)

(1)-(2)

(a-b)-(c-d)=sn-tn

(a-b)-(c-d)=n(s-t)

Both s and t are integers so s−t is also integer

put s−t=u

(a-b)-(c-d)=nu

By defination of Congruence

\(\displaystyle{\left({a}-{b}\right)}\equiv{\left({c}-{d}\right)}{\left(\text{mod}{n}\right)}\)

Hence Proved,

\(\displaystyle{\left({a}-{c}\right)}\equiv{\left({c}-{d}\right)}{\left(\text{mod}{n}\right)}\)