$k \equiv 3\ \ (mod\ 2) \Longleftrightarrow 2 | (k - 3)$
From
(10) we know that declaring $k$ congruent to $3$ mod $2$ is equivalent to declaring that $2$ divides $k - 3$.
(11)
$2 | (k - 3) \Longleftrightarrow \exists (m \in \mathbb{Z}) \land (k - 3 = 2m)$
From
(3) we know that declaring $2 | (k - 3)$ is equivalent to declaring that there exists some integer $m$ such that $k - 3 = 2m$
(12)
$k \equiv 3\ \ (mod\ 2) \Longleftrightarrow \exists (m \in \mathbb{Z}) \land (k - 3 = 2m)$
Let's just put (11) and (12) together and put a pin in it. (11) and (12) form a syllogism with (13)
(13)
Would you please pause and reflect on 13 for a moment?
$k \equiv 7\ \ (mod\ 2) \Longleftrightarrow 2 | (k - 7) \Longleftrightarrow \exists (m \in \mathbb{Z}) \land (k - 7 = 2m)$
Similar to what we found in (11), (12) and (13), we know that declaring $k$ congruent to $7$ mod $2$ is equivalent to declaring that $2$ divides $k-7$, which is equivalent to declaring that there exists some integer $m$ such that $k - 7 = 2m$
(14)
$(k - 7 = 2l) \Longleftrightarrow (k - 3 - 4 = 2l)$
Since $-7 = -3 - 4$, we can rewrite $(k - 7 = 2l)$ from (14) as $(k - 3 - 4 = 2l)$
(15)
$(k - 3 - 4 = 2l) \Longleftrightarrow (k - 3 = 2l + 4)$
By adding $4$ to both sides of the equality in (15), we get an equivalent expression $(k - 3 = 2l + 4)$
(16)
$(k - 3 = 2l + 4) \Longleftrightarrow (k - 3 = 2l + 2\cdot 2)$
By observing that $4 = 2 \cdot 2$ we may substitute $2 \cdot 2$ for $4$ in (16)
(17)
$(k - 3 = 2l + 2\cdot 2) \Longleftrightarrow (k - 3 = 2(l + 2))$
By factoring out 2 we may rewrite (17) as $(k - 3 = 2(l + 2))$
(18)
$(k - 3 = 2(l + 2)) \Longleftrightarrow (m := l + 2) \land (k - 3 = 2m)$
But hold on. Isn't $l + 2$ just another integer? Why can't we just define a new integer variable $m = l + 2$ to take the place of $l + 2$?
(19)
$(k - 3 = 2m) \Longleftrightarrow (k \equiv 3\ \ mod\ (2))$
But hold on again. Doesn't $k - 3 = 2m$ look familiar? From (13), isn't it equivalent to $(k \equiv 3\ \ mod\ (2))$? Yes, it is. □
(20)
