# $x\wedge a \wedge b=x \Rightarrow x\wedge b=x$ in a lattice viewed as an algebra

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

• 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”. – Arturo Magidin Aug 4 '19 at 22:37
• P.S. You forgot that $\wedge$ and $\vee$ are also idempotent. – Arturo Magidin Aug 4 '19 at 22:41
• Doesn't idempotency follow from the other axioms? @ArturoMagidin – why Aug 4 '19 at 23:21
• 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. – Arturo Magidin Aug 4 '19 at 23:25
• @ArturoMagidin thank you for clarifying! – why Aug 4 '19 at 23:27

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*}