I have the following question: Prove $a≥b$ if and only if $a+c≥b+c$ $\forall$ natural numbers. (using Peano Axioms)
The solution I checked is different from what I used and same applies to the ones I saw here (Proof that addition preserves order and Prove that addition preserves order. (for natural numbers) and some more)
I did the following:
Assuming this isn't true, then we have $a≥b$ and $a+c<b+c$
By cancellation law we get $a<b$ in the the second statement, which contradicts the first, end of proof.
Note: Cancellation law was stated in the book for $a+c=b+c$ but I think I can use is this way as well.
Alternative way I thought about to avoid not using the cancellation exactly as stated:
Assuming the statement isn't true then we have $a \geq b$ AND $b+c>a+c$. Thus, we have $b+c=a+c+k$ where $k$ is a positive ($>0$) natural number. This statement can be simplified to $b=a+k$ according to the cancellation law.
Plugging this in our main equation we get $a+k \geq a$ which is always wrong since we have $k>0$, thus we have a contradiction, end of proof.
Is my solution correct? If not, what's the flaw in it?