Please, I need help to prove the following result. I tried to write a proof, but I think that it's full of bugs
Theorem
Let $P,Q$ be two (complete) lattices. Then $P\oplus Q$ is a (complete) lattice.
My idea so far
Let $a,b\in P\oplus Q$, I would like to show that $a\vee b, a\wedge b\in P\oplus Q$.
Definition (Linear sum): Let $P,Q$ be two (disjoint) ordered sets. $P\oplus Q$ is $P\cup Q$ (disjoint union) with the following order relation:
$$x\le y\mbox{ in } P\oplus Q \iff\begin{matrix} x\le y \mbox{ in }P, \mbox{ if }x,y\in P \mbox{ or }\\ x\le y \mbox{ in }Q, \mbox{ if }x,y\in Q\mbox{ or}\\ x\in P \mbox{ and } y\in Q.\end{matrix}$$
Let $a,b\in P\oplus Q$.
Case 1. If $a,b\in P$, then $a\vee b,a\wedge b\in P$, so these elements belong to $P\oplus Q.$
Case 2. If $a,b\in Q$, then $a\vee b,a\wedge b\in Q$, so these elements belong to $P\oplus Q.$
Case 3. If $a\in P$ and $b\in Q$, then $a\le b$ in $P\oplus Q.$ Let us consider the set
$$U=\{a,b\}^u=\{x\in P\oplus Q\ |\ a\le x\mbox{ and }b\le x\}$$
It is not empty because $b$ belongs to it. Moreover $b$ is the least element of U, so $a\vee b=b$. Since $b\in Q$, then $a\vee b\in P\oplus Q$.
Let us consider the set
$$L=\{a,b\}^l=\{x\in P\oplus Q\ | \ x\le a,x\le b\}$$
It is not empty because $a\in L$. Moreover $a$ is the biggest element of L, so $a\wedge b=a\in P\implies a\wedge b\in P\oplus Q.$
I think I can conclude that $P\oplus Q$ is a lattice. Is this a correct proof? If not, could you help me to fix it? Thanks.
Regarding completeness, I still have to think about it. Any suggestion is welcome.
In case 3, you don't have to go through all that. If $a\in P$ and $b \in Q$, then, as you deduced, $a \leq b$, and it immediately follows that $a \wedge b = a$ and $a \vee b = b$.
Regarding completeness, if $P$ and $Q$ are complete, take $A \subseteq P \cup Q$, and you need to show that $\bigvee A \in P \oplus Q$ (I suppose you are aware that if this is true for any such $A$, then $P \oplus Q$ is also closed under arbitrary meets).
Now there are two cases, if $A \subseteq P$, then $\bigvee A \in P$ (it's the same in $P \oplus Q$ as in $P$); otherwise, taking $A' = A \cap Q$, we have $\bigvee A = \bigvee A'$ (calculated in $Q$).