The only zeros of zeta on the nonpositive real axis are the negative even integers, and every other zero lies in the open critical strip

Let $m\ge 1$. Substituting $s=-2m$ into [L1], the sine factor vanishes, $\mathrm{\Gamma }(1+2m)$ is finite by [L3], and $\zeta (1+2m)\mathrm{\ne }0$ by [L2]. Hence $\zeta (-2m)=0$.

For $x<0$ real and not a negative even integer, the sine factor in [L1] is nonzero. Also $1-x>1$, so [L2] gives $\zeta (1-x)\mathrm{\ne }0$, and [L3] with [L4] gives $\mathrm{\Gamma }(1-x)\mathrm{\ne }0$. Therefore [L1] forces $\zeta (x)\mathrm{\ne }0$. To handle $x=0$, let $s\to 0$ in [L1]: one has $\sin (\pi s\mathrm{/}2)\sim \pi s\mathrm{/}2$, $\mathrm{\Gamma }(1-s)\to 1$, and zeta has a simple residue-one pole at $1$, so $\zeta (1-s)\sim -1\mathrm{/}s$. Thus $\zeta (0)=-1\mathrm{/}2\mathrm{\ne }0$. Hence the only nonpositive real zeros are the numbers $-2,-4,\dots$.

Now let $\rho$ be any zero of zeta that is not one of the negative even integers. If $\mathrm{Re}\rho \le 0$, then [L2] gives $\zeta (1-\rho )\mathrm{\ne }0$ because $\mathrm{Re}(1-\rho )\ge 1$, and [L4] gives $\mathrm{\Gamma }(1-\rho )\mathrm{\ne }0$ unless $1-\rho$ is a nonpositive integer, which cannot happen when $\mathrm{Re}\rho \le 0$. Since the sine factor in [L1] vanishes only at even integers, step 2.1 rules out that possibility. Therefore [L1] cannot vanish at $\rho$, a contradiction. So every zero not listed in step 1.1 satisfies $\mathrm{Re}\rho >0$. Applying [L2] again excludes $\mathrm{Re}\rho \ge 1$, so every remaining zero lies in $0<\mathrm{Re}\rho <1$.

If $\rho$ is a nontrivial zero, then step 3.1 places it in the open critical strip. The sine and Gamma factors in [L1] are therefore finite and nonzero, so the functional equation gives $\zeta (1-\rho )=0$. On $\mathrm{Re}s>1$, the Dirichlet series in [L5] satisfies $\overline{\zeta (s)}=\zeta (\bar{s})$ term by term. Meromorphic continuation therefore extends this identity to all $s\mathrm{\ne }1$, so $\zeta (\rho )=0$ implies $\zeta (\bar{\rho })=0$.