\begin{definition}[Definition:Bounded Lattice/Definition 2]
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $\vee$ and $\wedge$ have identity elements $\bot$ and $\top$ respectively.
Then $\struct {S, \vee, \wedge, \preceq}$ is a '''bounded lattice'''.
Thus $\struct {S, \vee, \wedge, \preceq}$ is a '''bounded lattice''' {{iff}} the following axioms are satisfied:
{{begin-axiom}}
{{axiom|n = L0
       |lc= Closure
       |q = \forall a, b \in S
       |m = a \vee b \in S
       |mm= a \wedge b \in S
}}
{{axiom|n = L1
       |lc= Commutativity
       |q = \forall a, b \in S
       |m = a \vee b = b \vee a
       |mm= a \wedge b = b \wedge a
}}
{{axiom|n = L2
       |lc= Associativity
       |q = \forall a, b, c \in S
       |m = a \vee \paren {b \vee c} = \paren {a \vee b} \vee c
       |mm= a \wedge \paren {b \wedge c} = \paren {a \wedge b} \wedge c
}}
{{axiom|n = L3
       |lc= Idempotence
       |q = \forall a \in S
       |m = a \vee a = a
       |mm= a \wedge a = a
}}
{{axiom|n = L4
       |lc= Absorption
       |q = \forall a, b \in S
       |m = a \vee \paren {a \wedge b} = a
       |mm= a \wedge \paren {a \vee b} = a
}}
{{axiom|n = L5
       |lc= Identity elements
       |q = \exists \top, \bot \in S: \forall a \in S
       |m = a \vee \bot = a = \bot \vee a
       |mm= a \wedge \top = a = \top \wedge a
}}
{{end-axiom}}
Category:Definitions/Bounded Lattices
\end{definition}