Number theory........

prove if a is congruent to b mod n, and c is congruent to d mod n
then a+b is congruent to c+d mod n

The conclusion should be that a + c is congruent to b + d mod n. Since a is congruent to b mod n, n divides a - b. So there is an integer d such that a - b = nd. Similarly, since c is congruent to d mod n, there is an integer e such that c - d = ne. Thus (a + c) - (b + d) = (a - b) + (c - d) = nd + ne = n(d + e) = nq, where q is the integer d + e. Thus n divides (a + c) - (b + d), i.e., a + c is congruent to b + d mod n.

x = y mod z if x-y = nz = a multiple of z.

What you wrote isn't correct, because you wrote it wrong.

Let a=2 and b = 9
2 = 9 mod 7

Let c = 3 and d = 10
3 = 10 mod 7
but
a+b =
2+9 = 11 = 4 mod 7
and
c+d=
3+10 = 13 = 6 mod 7
so
a+b is NOT congruent to c+d

What I think you meant to write is
a+c = b+d mod n

a = b mod n so a-b = kn
c = d mod n so c-d = mn
for some integers k and m

Add those
a - b + c - d = kn + mn
rearrange
(a+c) - (b + d) = (k+m)n
That the definition for
a+c = b+d mod n, which is what I think you are trying to prove.

a is congruent to b mod n ==> a = b mod n ==> a = kn + b (k:integer)
c is congruent to d mod n ==> c = d mod n ==> c = qn+d (q:integer)

a+c = n(k+q) + (b+d) = nr + (b+d) (with r = k+q)
that means : a+c = (b+d) mod n ........proved
++++++++++++++++