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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?

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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 $x<y$.

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  • $\begingroup$ 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? $\endgroup$ – Spectacles Dec 25 '17 at 22:57
  • $\begingroup$ Yes, if two statements are both true, then they are equivalent. $\endgroup$ – Eric Wofsey Dec 25 '17 at 23:09
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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, $x+y+z<2y+z$

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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 $B$.

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