TITLE: How can we prove this Isomorphism?
QUESTION [5 upvotes]: How can we prove that $X$ is isomorphic to $Y$?
Note: all rows and columns are exact and diagram is commutative.

If we can do the following  transformation such that $\operatorname{H}(H'‎\rightarrow‎ I'‎\rightarrow‎ J)=X$ and $\operatorname{H}(A‎\rightarrow‎ B'‎\rightarrow‎ C')=Y$ then we have 
$$0=\operatorname{H}(D‎\rightarrow‎ E‎\rightarrow‎ F)\rightarrow‎‎ \operatorname{H}(H'‎\rightarrow‎ I'‎\rightarrow‎ J)‎\rightarrow‎‎ \operatorname{H}(A‎\rightarrow‎ B'‎\rightarrow‎ C')‎\rightarrow \operatorname{H}(E‎\rightarrow‎ F‎\rightarrow‎ G)=0‎$$
as required.

REPLY [2 votes]: In favor of this article (which should be well known to everyone in my opinion): The snake lemma by J. Wise, there is no need to take elements for diagram chasing; and the same technique applies to this diagram.
[EDIT: I finally found the time to create the diagrams.]
$X,A,B,C,D,E$ and $F$ (and $G$ via the identity map) have maps to $G$ and we can replace them by 
their kernels. By exactness, this doesn't effect $X$, $A,D$ and $E$ and gives a commutative diagram with exact columns and rows

where $C' = \ker(C\to G)$, $B' = \ker(B\to G)$ and $F' = \ker(F\to G)$.
Now $A$ has maps to $B', E, F'$ (and trivial maps to $I$, $C'$ and $J$). So if we take cokernels and denote these by double-primes $(\cdot)''$ we get the commutative diagram with exact columns and rows
.
Repeating the Arguments for the lower rectangle, i.e. taking kernels to $J$ and cokernels from $D$ we get a staircase

where all the non-trivial maps are isomorphisms. Hence $X$ is isomorphic to $Y$.