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a == b (mod 5) <==> EXISTS j. (a - b) = 5j by definition of congruence.
x == y (mod 5) <==> EXISTS k. (x - y) = 5k by definition of congruence.
a + x == b + y (mod 5) <==> EXISTS n. (a + x - b - y) = 5n by definition of congruence.
EXISTS n. (a + x - b - y) = 5n
EXISTS n. (a - b) + (x - y) = 5n
EXISTS n. (a - b) + x - y) = 5n
EXISTS n. 5j + 5k = 5n
EXISTS n. j + k = n
Is my proof true?
# We got a == b mod 5;
# As far as I know, we can represent a as a 5q1+r1, where r1 is b, and if we substract b from a, we got something divisible by 5*q1, where q1 is positive integer such as "n>=1"
# same with x==y mod 5, x --> 5q2+r2 // r2 is y
# if we take our problem, we and substitution,
we got (5q1+r1) + (5q2+r2) == (r1+r2) mod 5
# know if we are substracting first from one, we
should got something divisible by 5, without remainder.
# (5q1+r1)+(5q2+r2)-(r1+r2) = 5q1+r1+5q1+r2-r1-r2 = 5q1+5q2 = 5 * (q1 + q2), or we can say 5 * k, where k is (q1+q2), and this one should be divisible by 5, 'coz we multiple integer by 5.
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