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? 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.
 A: $$\vdash\exists x\in\mathbb Z\left(\exists j\in\mathbb Z\left(x=2j\right)\wedge\exists k\in\mathbb Z\left(x=2k+1\right)\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right)$$
$$\vdash\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)$$
$$\vdash\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)\rightarrow\exists x\in\mathbb Z\left(2x=1\right)$$
and
$$\vdash\neg\exists x\in\mathbb Z\left(2x=1\right)$$
by repeated modus tollens
$$\vdash\neg\exists x\in\mathbb Z(\exists j\in\mathbb Z(x=2j)\wedge\exists k\in\mathbb Z(x=2k+1))$$
A: $\def\fitch#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}}$ 
$\fitch{
}{
1. \exists x\in\mathbb Z\left(\exists j\in\mathbb Z\left(x=2j\right)\wedge\exists k\in\mathbb Z\left(x=2k+1\right)\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right) \text{Theorem}\\
2. \exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right)\rightarrow\exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)\text{Theorem}\\
3. \exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)\rightarrow\exists x\in\mathbb Z\left(2x=1\right)\text{Theorem}\\
4. \neg \exists x\in\mathbb Z\left(2x=1\right)\text{Theorem}\\
\fitch{
5. \exists x\in\mathbb Z\left(\exists j\in\mathbb Z\left(x=2j\right)\wedge\exists k\in\mathbb Z\left(x=2k+1\right)\right) \quad \text{ Assumption}
}{
6. \exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2k+1=2j\right) \quad \rightarrow \text{ Elim } 1,5\\
7. \exists j\in\mathbb Z,\exists k\in\mathbb Z\left(2\left(j-k\right)=1\right)\quad \rightarrow \text{ Elim } 2,6\\
8. \exists x\in\mathbb Z\left(2x=1\right)\quad \rightarrow \text{ Elim } 3,7\\
9. \bot\quad \bot \text{ Intro } 4,8} \\ 
10. \neg \exists x\in\mathbb Z\left(\exists j\in\mathbb Z\left(x=2j\right)\wedge\exists k\in\mathbb Z\left(x=2k+1\right)\right) \quad \neg \text{ Intro } 5-9
}$
