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 $A$


Statement $B$


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 $A$?


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.


$B$ 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 $x \in Z$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.