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A lattice is an algebraic structure $(L,\wedge,\vee)$ such that, $\wedge$ and $\vee$ are commutative, associative and abosrbing binary operations, i.e.

$$a \wedge (b\vee a)=a,\quad a\vee(a\wedge b)=a.$$

I want to show $$x\wedge a \wedge b=x \Rightarrow x\wedge b=x.$$

Working from a lattice as a poset, and defining $a\wedge b$ as the essential source of $\{a,b\}$, i.e $$x=a\wedge b \iff (x\leq a) \quad (x\leq b) \quad \forall y\in L (y\leq a \text{ and } y\leq b)\Rightarrow y\leq x.$$ Then the result follows. But how do I show this without a relation

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  • $\begingroup$ Note: “algebraic lattice” has a different meaning: it is a complete lattice in which every element is a (possibly infinite) join of compact elements. It does not mean “lattice viewed as an algebra”. $\endgroup$ – Arturo Magidin Aug 4 '19 at 22:37
  • $\begingroup$ P.S. You forgot that $\wedge$ and $\vee$ are also idempotent. $\endgroup$ – Arturo Magidin Aug 4 '19 at 22:41
  • $\begingroup$ Doesn't idempotency follow from the other axioms? @ArturoMagidin $\endgroup$ – why Aug 4 '19 at 23:21
  • $\begingroup$ They can be derived using the absorption laws, but one usually includes them so that you can define upper subsemilattice and lower subsemilattices by taking a subset of axioms. $\endgroup$ – Arturo Magidin Aug 4 '19 at 23:25
  • $\begingroup$ @ArturoMagidin thank you for clarifying! $\endgroup$ – why Aug 4 '19 at 23:27
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If $x\wedge a \wedge b = x$, then $$\begin{align*} x\wedge b &= (x\wedge a\wedge b) \wedge b &\text{(substitution)}\\ &= (x\wedge a)\wedge (b\wedge b)&\text{(associativity)}\\ &= (x\wedge a) \wedge b&\text{(idempotency)}\\ &= x\wedge a \wedge b & \text{(associativity)}\\ &= x. &\text{(substitution)} \end{align*} $$

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