Top is Meet Irreducible

Theorem
Let $\left({S, \wedge, \preceq}\right)$ be a bounded above meet semilattice.

Then $\top$ is meet irreducible

where $\top$ denotes the greatest element in $S$.

Proof
Let $x, y \in S$ such that
 * $\top = x \wedge y$

By Meet Precedes Operands
 * $\top \preceq x$ and $\top \preceq y$

By definition of greatest element:
 * $x \preceq \top$

Thus by definition of antisymmetry:
 * $\top = x$