Proof by contradiction: $c<a<d \wedge c<b<d \to |a-b|<d-c$

Question

Let be $a,b,c,d \in \mathbb{R}$, I must proof "$c<a<d \wedge c<b<d \to |a-b|<d-c$".

Proof by contradiction: I have $|a-b|\geq d-c$, therefore $a-b \leq c-d \vee a-b \geq d-c$ (or $a-c \leq b-d \vee a-d \geq b-c$, but by hypothesis I have $c<a<d \wedge c<b<d$ then $a-c >0$,$b-d<0$,$a-d<0$ and $b-c>0$ therefore $a-c \leq b-d$ and $a-d \geq b-c$ are absurd. Is correct?

Thanks in advance!

P.S.= "$c\leq a \leq d \wedge c \leq b \leq d \to |a-b| \leq d-c$" is true?

Answer 1

That's all correct as far as I can see. The statement is also true in the non-strict inequality case.

Answer 2

And, contradiction is not needed here. Just split into the cases $a \le b$ and $b \le a$ to remove the absolute sign, and you can prove the inequality directly.