$\forall x\in\mathbb{N}[(\exists y\in\mathbb{N}(2\leq y\land y<x\land y\ |\ x))\to(\exists z\in\mathbb{N}(2\leq z\land z\leq\sqrt x\land z\ |\ x))]$

Question

Please help me to prove this statement. I'm very confused, Thank you! I Have tried prove it using examples but it needs to be proved using mathematical method

Answer 1

Note: Remember our assumption that $\exists y \ge 2 \space \& \space y|x$ means $x$ must be non-prime number greater than $4$.

We have $2$ cases, the trivial case where $y \le \root \of x$ in which case $z = y$ and the other case $y > \root \of x$

Let $y > \root \of x\space \& \space y<x$, and $y|x \implies \exists z \in \mathbb N$ such that $yz = x$. Since $y > \root \of x \implies z < \root \of x$ (This is fairly easy to prove). In addition, $yz = x \implies z|x$. Since $y<x \implies z>1 \implies z\ge 2$.

Therefore $∀x \in \mathbb N$ such that$ [(∃y \in \mathbb N (2 ≤ y ∧ y < x ∧ y | x)) → (∃z \in \mathbb N(2 ≤ z ∧ z ≤ √x ∧ z | x))]$

Answer 2

Feeding the negation into a tableau gives

,

which is closed apart from the $\neg(b\le \sqrt{a})$ path.