Proving is $x+1$ is even, then $x$ is odd

Question

Suppose that $x+1$ is even, such that there exists and integer $k$ such that $x+1=2k$. $$x+1 = 2k\implies x=2k-1$$ since $k$ is an integer and $L+1$ is also an integer $k := L+1$
$$x=2(L+1)-1 \implies x=2L+2-1 \implies x=2L+1$$
Since $x$ follows the definition of odds, if $x+1$ is even, $x$ is odd.

Is this proof sound? I feel like I might have gone awry when I said "since $k$ is an integer and $L+1$ is also an integer $k := L+1$."

Answer 1

It would be better to say: "Since $k$ is an integer, then $l:=k-1$ is an integer, and $k=l+1$..." then show that $x=2l+1$. Aside from that, it looks just fine!