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Let us define an algebraic lattice as a triple $(A,\wedge,\vee)$ in which $\wedge$ and $\vee$ are associative and commutative binary operators which satisfy the laws of absorption: $x{\vee}(x{\wedge}y)=x=x{\wedge}(x{\vee}y)$.

Let us take for granted that we have shown somehow that for $\le$ defined via $x{\le}y\iff x{\wedge}y=x$ the pair $(A,\le)$ is a poset lattice, i.e. it is a poset in which every pair of elements has a supremum and an infimum. I would like to show in addition that the supremum coincides with $\vee$.

Let us start:

  • $x\le x{\vee}y$ since $x{\wedge}(x{\vee}y)=x$.
  • $y\le x{\vee}y$ since $y{\wedge}(x{\vee}y)=y{\wedge}(y{\vee}x)=y$.

So $x{\vee}y$ is an upper bound of $x$ and $y$. To show that $x{\vee}y$ is the least upper bound of $x$ and $y$, let $z$ be any upper bound of $x$ and $y$. Then $x{\wedge}z=x$ and $y{\wedge}z=y$. We need to show that $(x{\vee}y){\wedge}z$ is equal to $x{\vee}y$.

How? Prove or give a counterexample.

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    $\begingroup$ You should replace "algebraic lattice" with something like "algebraic definition of lattice" or similar. The reason is that there is a notion of algebraic lattice (en.wikipedia.org/wiki/Compact_element#Algebraic_posets). So while the result you're seeking is true for algebraic lattices, it is also true for any lattice. $\endgroup$
    – amrsa
    Aug 21, 2016 at 11:47
  • $\begingroup$ Nothing occurs to me. As far as I ever read about, they're just called lattices, one way or the other. In fact, we're talking about different presentations of the same objects... $\endgroup$
    – amrsa
    Aug 22, 2016 at 9:04

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$$x=x\wedge z$$so by absorption$$x\vee z=(x\wedge z)\vee z = z$$and similirly$$y\vee z= z$$ so$$ (x\vee y)\wedge z= (x\vee y)\wedge (x\vee z) = (x\vee y)\wedge (x\vee(y\vee z)) = (x\vee y)\wedge ((x\vee y)\vee z))$$and again, by absorption $$=x\vee y$$

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  • $\begingroup$ A short-cut: if $z$ is an upper bound of $x$ (and $y$), then $x \vee z = z$ (no need for using the absortion law) $\endgroup$
    – amrsa
    Aug 21, 2016 at 11:42
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    $\begingroup$ @LeonMeier i remember somthing like this in Gratzer's "General lattice theory" $\endgroup$
    – user160823
    Aug 21, 2016 at 18:17
  • $\begingroup$ @LeonMeier "Not directly,..." Good point. $\endgroup$
    – amrsa
    Aug 22, 2016 at 9:05
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    $\begingroup$ @LeonMeier i looked it up. he does the dual proof in the bottom of page 5 in the book. $\endgroup$
    – user160823
    Aug 22, 2016 at 12:33

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