Chinese Remainder Theorem, redundant information I want to solve the following system of congruences:
$ x \equiv 1 \mod 2 $
$ x \equiv 2 \mod 3 $
$ x \equiv 3 \mod 4 $
$ x \equiv 4 \mod 5 $
$ x \equiv 5 \mod 6 $
$ x \equiv 0 \mod 7 $
I know, but do not understand why, that the first two congruences are redundant. Why is this the case? I see that the modulo of the congruences are not pairwise relatively prime, but why does this cause a redundancy or contradiction? Further, why is it that in the solution to this system, we discard the first two congruences and not 
$ x \equiv 3 \mod 4 $
$ x \equiv 5 \mod 6 $
being that $ gcd(3,6) = 3 $ and $gcd(2,4) = 2$ ?
EDIT:
How is the modulo of the unique solution effected if I instead consider the system of congruences without the redundancy i.e. does $M = 4 * 5 * 6 * 7$ or does it remain $M= 2*3*4*5*6*7$?
 A: Note that:
$$x\equiv 3 \mod 4 \Rightarrow x= 4k+3\Rightarrow x\equiv 1 \mod 2\\ x\equiv 5\mod 6 \Rightarrow x = 6k'+5\Rightarrow x\equiv 2\mod 3
$$
A: Hint $\ x\equiv -1\ $ mod $\,2,3,4,5,6\iff x\equiv -1 \pmod m\ $ for $\, m = {\rm lcm}(2,3,4,5,6) = {\rm lcm}(4,5,6)$
because $\ 2,3,4,5,6\mid x\!+\!1\iff 4,5,6\mid x\!+\!1,\ $ since $\,4,6\mid x\!+\!1\,\Rightarrow\,2,3\mid x\!+\!1$
A: $x \equiv 1 \mod 2$
$x \equiv 2 \mod 3$
$x \equiv 3 \mod 4 \implies x \equiv 1 \pmod 2$
$x \equiv 4 \mod 5$
$x \equiv 5 \mod 6 \iff
\left.\begin{cases}
   x \equiv 1 \mod 2 \\
   x \equiv 2 \mod 3
\end{cases} \right\}$
By the CRT.
$x \equiv 0 \mod 7$

So first we replace $x \equiv 5 \mod 6$ with 
$\left.\begin{cases}
   x \equiv 1 \mod 2 \\
   x \equiv 2 \mod 3
\end{cases} \right\}$

$x \equiv 1 \mod 2$
$x \equiv 2 \mod 3$
$x \equiv 3 \mod 4 \implies x \equiv 1 \pmod 2$
$x \equiv 4 \mod 5$
$x \equiv 1 \mod 2$
$x \equiv 2 \mod 3$
$x \equiv 0 \mod 7$

Now, because $x \equiv 1 \pmod 2$ is redundant, we remove all instanced of it and we remove all but one instance of $x \equiv 2 \mod 3$.

$x \equiv 2 \mod 3$
$x \equiv 3 \mod 4$
$x \equiv 4 \mod 5$
$x \equiv 0 \mod 7$

In this case, we can cheat a little if we change the first three congruences to equivalent congruences.
$
\left. 
\begin{array}{l}
   x \equiv -1 \mod 3 \\
   x \equiv -1 \mod 4 \\
   x \equiv -1 \mod 5
\end{array}
\right\} \ \iff x \equiv -1 \mod 60$
(Again, by the CRT.)
$x \equiv 0 \mod 7$

So we now have

$x \equiv -1 \mod 60$
$x \equiv 0 \mod 7$

To solve this, we start with $x \equiv 0 \mod 7$, which implies that
$x = 7n$ for some integer $n$. Substitute that into $x \equiv -1 \mod 60$ and you get
$7n \equiv -1 \mod 60$
So we need to find $\dfrac 17 \pmod{60}$ The most fundamental way to do this is to inspect numbers congruent to $1 \pmod{60}$ until we find one that is a multiple of $7$. At worst, we will have to examint $7$ such numbers.
$1, 61, 121, 181, 241, \color{red}{301}$
Since $7 \times 43 = 301$, then $\dfrac 17 \equiv 43 \pmod{60}$. So we conclude that $n \equiv -43 \equiv 17 \mod{60}$.
Then $x = 7n = 119 \mod{420}$.
