Good proof / Bad proof

We know that zero is even, since $0 = 2*0$. Could it also be odd? Below are three purported proofs that zero is, in fact, not odd. One of these proofs is a correct proof, but the other two are not. Obviously the thing they are trying to prove is true, but the proofs have flaws.

Your job: spot the correct proof, and figure out what's wrong with each of the incorrect proofs.

Zero is not odd. Suppose by way of contradiction that 0 is odd. Then there is an integer $k$, such that $2*k+1 = 0$. But adding $-1$ to both sides, we get $2*k = -1$. This means that $-1$ is even. However, since $-1 = 2*(-1) + 1$, $-1$ is odd, so we have a contradiction. Therefore $0$ is not odd.

Zero is not odd. Assume by way of contradiction that odd(0). Then let $k$ be an integer such that $2*k+1=0$. By the trichotemy law, there are three cases: case $k = 0$: then $2*k+1 = 1 \neq 0$. case $0 \lt k$: then $2*0 \lt 2*k$ and so $0 \lt 2*k$. Since $-1 \lt 0$, by the transitivity of $\lt$, $-1 \lt 2*k$. Adding 1 to both sides gives $0 \lt 2*k + 1$, so $2*k+1 \neq 0$. case $k \lt 0$: then $k = -1$ or $k \lt -1$ by the Gaps everywhere theorem. if $k = -1$ then $2*k + 1 = -2 + 1 = -1 \neq 0$ if $k \lt -1$ then $2*k \lt -2$ and so $2*k + 1 \lt -1$, therefore $2*k+1 \neq 0$ In all cases $2*k+1 \neq 0$, which contradicts our assumption. So there is no $k$ such that $2*k+1 = 0$, i.e. $\neg \exists k[0 = 2*k+1]$, therefore $\neg \text{odd}(0)$.

Zero is not odd. If zero is odd, then there needs to be a number $k$ such that $2*k+1 = 0$, so we will use induction to prove that for any $k$, $2*k+1 \neq 0$. case $k=0$: then $2*k+1 = 1 \neq 0$. case $0 \lt k$: then $2*0 \lt 2*k$ and so $0 \lt 2*k$. Since $-1 \lt 0$, by the transitivity of $\lt$, $-1 \lt 2*k$. Adding 1 to both sides gives $0 \lt 2*k + 1$, so $2*k+1 \neq 0$. So by induction we deduce that there is no integer such that $2*k+1 \neq 0$, which means that 0 is not odd.