# Am I allowed to do this to prove that one statement proves the other?

This is sort of a continuation of my previous post but different statements: Does this prove one statement implies the other?

Statement $A$

$∀a,b,c∈ℤ,a<b<c⇒a+b+c<2b+c$

Statement $B$

$∀x,y,z∈ℤ,x<y∧z≠x∧z≠y⇒x+y+z<2y+z$

I'd like to prove $A$ implies $B$ by saying that, assuming the antecedent of $B$,

$z<x<y∨x<z<y∨x<y<z$

Could I then say to take $a$ from $A$ to be min{$x,y,z$}, $c$ from $A$ to be max{$x,y,z$}, and $b$ to be the one left over showing:

$x+y+z<2y+z$?

Also if the proof is invalid but the implication is true, could you show me how to prove it?

Well, your method is valid, but your last step is wrong. You can't conclude $x+y+z<2y+z$; rather, you can conclude $a+b+c<2b+c$ for the definitions of $a,b,$ and $c$ you have made. Depending on which case you are in, this gives you a different inequality. For instance, in the case $z<x<y$ it gives you $$x+y+z<2x+y,$$ because $a=z$, $b=x$, and $c=y$ in that case.

To prove statement A implies statement B, I would instead suggest trying to prove statement B directly (after all, if statement B is true, then anything implies it). Given $x<y$, how could you obtain the inequality $x+y+z<2y+z$? The answer is hidden below.

Just add $y+z$ to both sides of $x<y$.

• When you said "if statement B is true, then anything implies it", does that mean that if I can prove both statements to be true, it shows that they are equivalent? – Spectacles Dec 25 '17 at 22:57
• Yes, if two statements are both true, then they are equivalent. – Eric Wofsey Dec 25 '17 at 23:09

While mathematically true, statement B does not logically follow from statement A. To prove statement you need to prove it for all integers $x$, $y$ and $s$ and not just the ones where $z$ is the greatest of those three. So, statement A by itself is not enough to imply statement B, since A is only about that restricted set.

Your particular proof does not work,because when $z<x<y$, all that you can get from statement A is that $z+x+y<2x+y$, which is not the same as, nor implies, $x+y+z<2y+z$

To add to what others have said in other answers: actually you can use $B$ to prove $A$. This is because, if $a\lt b\lt c$ then $a\lt b$ and $c\ne a$ and $c\ne b$, so you can 'plug' $x=a, y=b, z=c$ in $B$.