Weak axiom of revealed preference and choice coherence - how to show they are equivalent $B$ and $B'$ are elements of the family of subsets of $X$

WARP
For every pair $x,y \in B \cap B' $ and if $x \in c(B)$ , then if $y \in c(B'), x$ must $\in c(B').$

Choice Coherence
For very pair $x,y \in B \cap B'$ and if $x \in c(B)$ and $y  \notin c(B)$, then $y$ must $\notin c(B').$

Are these two equivalent. If yes, how can we prove it?
Yes, they are equivalent. Here is a formal proof by contradiction.

WARP $\Rightarrow$ Choice coherence

Suppose that WARP holds but that choice coherence is not true. There exists $B,B'$, $x,y \in B \cap B'$ such that $x \in c(B)$, $y \notin c(B)$ and $y \in c(B')$. 

But WARP applied to the conditions $y \in c(B')$ and $x \in c(B)$ implies $y \in c(B)$. This is a contradiction.

Choice coherence $\Rightarrow$ WARP

Suppose now that choice coherence is true but that WARP is falsified. There exists $B,B'$, $x,y \in B \cap B'$ such that $x \in c(B), y \in c(B')$ and $x \notin c(B')$. 

But choice coherence applied to the conditions $y \in c(B')$, $x \notin c(B')$ yields $x \notin c(B)$. This is a contradiction.