Theorem: Equivalence of Definitions of Lattice Isomorphism/Definition 2 Implies Definition 1

Let $L_1 = \struct {A_1, \vee_1, \wedge_1, \preceq_1}$ and $L_2 = \struct {A_2, \vee_2, \wedge_2, \preceq_2}$ be lattices.
Let $\phi : \struct{A_1, \preceq_1} \to \struct{A_2, \preceq_2}$ be an order isomorphism.
Then:
:$\phi$ is a bijective lattice homomorphism

Proof:
By definition of order isomorphism:
:$\phi : \struct{A_1, \preceq_1} \to \struct{A_2, \preceq_2}$ is an order-preserving bijection
Let $\phi^{-1} : A_2 \to A_1$ be the inverse of $\phi : A_1 \to A_2$.
From Inverse of Order Isomorphism is Order Isomorphism:
:$\phi^{-1} :  \struct{A_2, \preceq_2} \to \struct{A_1, \preceq_1}$ is an order isomorphism
==== $\phi$ Satisfies Join Morphism Property ====
Let:
:$x, y \in A_1$
We have:
{{begin-eqn}}
{{eqn | l = x, y
      | o = \le
      | r = x \vee_1 y
      | c = {{Defof|Supremum}}
}}
{{eqn | ll = \leadsto
      | l = \map \phi x, \map \phi y
      | o = \le
      | r = \map \phi {x \vee_1 y}
      | c = {{Defof|Order-Preserving Mapping}}
}}
{{eqn | ll = \leadsto
      | l = \map \phi x \vee_2 \map \phi y
      | o = \le
      | r = \map \phi {x \vee_1 y}
      | c = {{Defof|Supremum}}
}}
{{eqn | ll = \leadsto
      | l = \map {\phi^{-1} } {\map \phi x \vee_2 \map \phi y}
      | o = \le
      | r = \map {\phi^{-1} } {\map \phi {x \vee_1 y} }
      | c = {{Defof|Order-Preserving Mapping}}
}}
{{eqn | n = 1
      | ll = \leadsto
      | l = \map {\phi^{-1} } {\map \phi x \vee_2 \map \phi y}
      | o = \le
      | r = x \vee_1 y
      | c = {{Defof|Inverse Mapping}}
}}
{{end-eqn}}
Also we have:
{{begin-eqn}}
{{eqn | l = \map \phi x, \map \phi y
      | o = \le
      | r = \map \phi x \vee_2 \map \phi y
      | c = {{Defof|Supremum}}
}}
{{eqn | ll = \leadsto
      | l = \map {\phi^{-1} } {\map \phi x}, \map {\phi^{-1} } {\map \phi y}
      | o = \le
      | r = \map {\phi^{-1} } {\map \phi x \vee_2 \map \phi y}
      | c = {{Defof|Order-Preserving Mapping}}
}}
{{eqn | ll = \leadsto
      | l = x, y
      | o = \le
      | r = \map {\phi^{-1} } {\map \phi x \vee_2 \map \phi y}
      | c = {{Defof|Inverse Mapping}}
}}
{{eqn | n = 2
      | ll = \leadsto
      | l = x \vee_1 y
      | o = \le
      | r = \map {\phi^{-1} } {\map \phi x \vee_2 \map \phi y}
      | c = {{Defof|Supremum}}
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = x \vee_1 y
      | r = \map {\phi^{-1} } {\map \phi x \vee_2 \map \phi y}
      | c = {{Ordering-axiom|3}} applied to $(1)$ and $(2)$ 
}}
{{eqn | ll = \leadsto
      | l = \map \phi {x \vee_1 y}
      | r = \map \phi {\map {\phi^{-1} } {\map \phi x \vee_2 \map \phi y} }
}}
{{eqn | r = \map \phi x \vee_2 \map \phi y
      | c = {{Defof|Inverse Mapping}}
}}
{{end-eqn}}
It follows:
:$(3):\quad \forall x, y \in A_1 : \map \phi {x \vee_1 y} = \map \phi x \vee_2 \map \phi y$
That is, $\phi$ satisfies the join morphism property.
{{qed|lemma}}
==== $\phi$ Satisfies Meet Morphism Property ====
The dual statement of $(3)$ is:
:$(4):\quad \forall x, y \in A_1 : \map \phi {x \wedge_1 y} = \map \phi x \wedge_2 \map \phi y$
by Dual Pairs (Order Theory).
We have $(4)$ holds from Duality Principle.
That is, $\phi$ satisfies the meet morphism property. 
{{qed|lemma}}
It follows that $\phi$ is a bijective lattice homomorphism by definition.