TITLE: Proving Statement $A$ implies Statement $B$
QUESTION [5 upvotes]: I'm trying to wrap my head around logical implications, equivalences, etc. and I made up two random statements. This is my attempt at proving one implies the other:
Statement $A$
$∀a,b,c∈Z,a<b<c⇒0<f(a)<f(b)<f(c)$
Statement $B$
$∀x,y,z∈Z,x<y∧z≠x∧z≠y⇒f(x)+f(y)+f(z)>0$
Prove $A⇒B$
Assume Statement $A$ and the antecedent of Statement $B$
Dividing Statement $B$ into cases:
Case 1: $z<x<y$
From $A$ we can conlude that $0<f(z)<f(x)<f(y)$, meaning $f(x)+f(y)+f(z)>0$
Case 2: $x<z<y$ and Case 3: $x<y<z$
(Same reasoning as in case 1)
By showing that all three possible cases are true given the assumption that $A$ is true, can I say that Statement $A$ implies Statement $B$? 
And if possible, how would you prove that $B$ implies $A$?

REPLY [3 votes]: This is interesting.  Your argument that the truth of $A$ forces the truth of $B$ is correct.  You have divided it into sensible cases and correctly argued the cases.  As an exercise in logic or proof-writing, it's fine.
There is, however, quite a bit of fat that can be trimmed out of all this to get to the real meat of the problem.  
Your statement $A$ immediately implies that $f$ is a positive function, so that $f(x) > 0$ for all $x \in \mathbb{Z}$.  Take any $a$, and let $b=a+1$ and $c=a+2$.  Then you can conclude $f(a) > 0$ from $A$.  Since $a$ is arbitrary, we're done.  Now that $f$ is positive, $B$ follows immediately:  the sum of three positive integers is again positive.