How do you prove $\neg\exists x\in\mathbb Z(\exists j\in\mathbb Z(x=2j)\wedge\exists k\in\mathbb Z(x=2k+1))$ by contradiction and natural deduction?

Question

There are three implications which I think will be of help: $$\exists x\in\mathbb Z\left(\exists j\in\mathbb Z\left(x=2j\right)\wedge\exists k\in\mathbb Z\left(x=2k+1\right)\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right)$$ $$\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)$$ $$\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)\rightarrow\exists x\in\mathbb Z\left(2x=1\right)$$ And we know that $\exists x\in\mathbb Z\left(2x=1\right)$ is false.

Answer 1

$\def\fitch#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}}$

$\fitch{ }{ 1. \exists x\in\mathbb Z\left(\exists j\in\mathbb Z\left(x=2j\right)\wedge\exists k\in\mathbb Z\left(x=2k+1\right)\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right) \text{Theorem}\\ 2. \exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)\text{Theorem}\\ 3. \exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)\rightarrow\exists x\in\mathbb Z\left(2x=1\right)\text{Theorem}\\ 4. \neg \exists x\in\mathbb Z\left(2x=1\right)\text{Theorem}\\ \fitch{ 5. \exists x\in\mathbb Z\left(\exists j\in\mathbb Z\left(x=2j\right)\wedge\exists k\in\mathbb Z\left(x=2k+1\right)\right) \quad \text{ Assumption} }{ 6. \exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right) \quad \rightarrow \text{ Elim } 1,5\\ 7. \exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)\quad \rightarrow \text{ Elim } 2,6\\ 8. \exists x\in\mathbb Z\left(2x=1\right)\quad \rightarrow \text{ Elim } 3,7\\ 9. \bot\quad \bot \text{ Intro } 4,8} \\ 10. \neg \exists x\in\mathbb Z\left(\exists j\in\mathbb Z\left(x=2j\right)\wedge\exists k\in\mathbb Z\left(x=2k+1\right)\right) \quad \neg \text{ Intro } 5-9 }$

Answer 2

$$\vdash\exists x\in\mathbb Z\left(\exists j\in\mathbb Z\left(x=2j\right)\wedge\exists k\in\mathbb Z\left(x=2k+1\right)\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right)$$ $$\vdash\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)$$ $$\vdash\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)\rightarrow\exists x\in\mathbb Z\left(2x=1\right)$$ and $$\vdash\neg\exists x\in\mathbb Z\left(2x=1\right)$$ by repeated modus tollens $$\vdash\neg\exists x\in\mathbb Z(\exists j\in\mathbb Z(x=2j)\wedge\exists k\in\mathbb Z(x=2k+1))$$