If $x>y$, then $\lfloor x\rfloor\ge \lfloor y\rfloor$, formal proof For x ∈ ℝ, define by:       ⌊x⌋ ∈ ℤ ∧ ⌊x⌋ ≤ x ∧ (∀z ∈ ℤ, z ≤ x ⇒ z ≤ ⌊x⌋).
Claim 1.1:  ∀x  ∈ ℝ, ∀y  ∈ ℝ, x  > y  ⇒ ⌊x⌋ ≥ ⌊y⌋.
Assume, x, y ∈ ℝ # Domain assumption
Assume x  > y               # Antecedent assumption

    Then x ≥ ⌊x⌋            # By definition

    Then x ≥ z              # By definition

    Then ⌊x⌋≥ z             # Also by definition

    Let z = ⌊y⌋             # By assumption & definition since ∀z ∈ ℤ, y ∈ ℝ and ℤ ∈ ℝ

    Then ⌊x⌋ ≥ ⌊y⌋          # Substitution

Then x  > y  ⇒ ⌊x⌋ ≥ ⌊y⌋    # Introduce implication

Then ∀x  ∈ ℝ, ∀y  ∈ ℝ, x  > y  ⇒ ⌊x⌋ ≥ ⌊y⌋  # Introduce universal

$x\geq y$ combined with $\lfloor y\rfloor\leq y$ gives $\lfloor y\rfloor\leq x$.
Here $\lfloor y\rfloor\in\mathbb Z$ so we conclude that $\lfloor y\rfloor\leq \lfloor x\rfloor$.
Proved is now: $$x\geq y\Rightarrow \lfloor x\rfloor\geq \lfloor y\rfloor$$
