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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 $A$?

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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.

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$B$ does not imply $A$. $B$ is true if $f$ is any function that is always positive. $A$ is not true, e.g. if $f(x)=1$ for all $x \in Z$.

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  • $\begingroup$ But aren't I assuming statement $A$? I.e. $f$ satisfies $A$ and now I am trying to show that $f$ will satisfy $B$? If I have incorrectly depicted that scenario with this question, how should I word it to depict what I just mentioned? $\endgroup$ – Spectacles Dec 26 '17 at 1:22
  • $\begingroup$ @Spectacles Robert Israel is talking about proving $B \implies A$. In this case, no you are not assuming $A$. The OP gave you an instance of the statement $B$ being true at the same time as statement $A$ being false (namely the instance when $f(x)=1$ for all $x \in \mathbb{Z}$). Therefore $B \implies A$ is false. $\endgroup$ – Ovi Dec 26 '17 at 3:01

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