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I am having trouble with the following question

Show that a lattice is distributive iff for any element $a,b,c$ in the lattice $$(a\lor b)\land c \leq a \lor(b\lor c)$$

My attempt:

Let the lattice be distributive. Hence $$(a \lor b) \land c=(a \land c) \lor (b \land c) ~~~...\mathbf i)$$

We also have $$a \land c \leq a ~and~ b \land c \leq b \lor c $$ Combing the two we get, $$(a \land c) \lor (b \land c) \leq a \lor (b \lor c)$$ Using $\mathbf i)$ we can prove that $$(a\lor b)\land c \leq a \lor(b\lor c)$$ Now I do not know how to prove it the other way around. If anyone has a better way to prove it then please suggest it to me. Thanks in advance.

EDIT: Okay I found the question in a text book. It seems the question is $$(a\lor b)\land c \leq a \lor(b\land c) $$. I will try the problem once again and post if I face any difficulty.

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  • $\begingroup$ That doesn't seem right, because $(a\lor b)\land c \leq a \lor(b\lor c)$ is true for all lattices. Did the question maybe say: $$(a\lor b)\land c \leq a \lor(b\land c)$$ $\endgroup$ – Thomas Andrews Sep 5 '14 at 16:17
  • $\begingroup$ I was wondering the same thing. The given statement is always true. $\endgroup$ – GTX OC Sep 5 '14 at 16:29
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The condition you've written is true in all lattices, so, since non-distributive lattices exists, the theorem can't be true. Perhaps the question was:

$$(x\lor y)\land z \leq x \lor(y\land z)$$

It's easy to show: $$(a\lor b)\land c\geq (a\land c)\lor (b\land c)$$ in any lattice.

So you need to show that $$(a\lor b)\land c\leq (a\land c)\lor (b\land c)$$

Letting $x=(a\land c)$, $y=b$, $z=c$, we get from the above rule:

$$((a\land c)\lor b)\land c\leq (a\land c)\lor (b\land c)$$

But with $x=b,y=a,z=c$, we get: $$(b\lor a)\land c\leq b\lor(a\land c)=(a\land c)\lor b$$ Since $c\geq (b\lor a)\land c$, that means $$(a\lor b)\land c=(b\lor a)\land c\leq ((a\land c)\lor b)\land c\leq (a\land c)\lor (b\land c)$$ and we are done.

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