0
$\begingroup$

I am attempting to prove the equivalence of the following two definitions of distributive lattices:

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

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

I haven't figured it out yet, but along the way I came up with this probably bogus proof:

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

Renaming variables (rename b to c and c to b):

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

Rename with $b = b \lor c$ (by noticing the left side of the equation looks like the right side of the second condition):

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

Using the identities $b \land c = b \land c \land c$ and $c \lor (b \land c) = c$, simplify:

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

Now, here's the dodgy step. Once again, rename $b \lor c$ to $b$, which concludes the proof:

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

Where does this proof go wrong (or is it fine?), and what is the proof strategy for properly showing this equivalence?

Improve this question
$\endgroup$
2
  • $\begingroup$ I didn't read your thing in detail, but look at pages 1-2 of math.hawaii.edu/~jb/math618/os8uh.pdf The absorption laws for lattices are invoked twice; see en.wikipedia.org/wiki/… By the way, this material is very standard and not appropriate for MO. You might get a more detailed response at math.SE. $\endgroup$
    – Todd Trimble
    Commented Mar 20, 2011 at 16:11
  • $\begingroup$ Where you wrote "Rename $b$ with $b\vee c$," your next equation changed the last two occurrences of $b$ to $b\vee c$ as promised but changed the first occurrence to $b\wedge c$. $\endgroup$
    – Andreas Blass
    Commented Mar 20, 2011 at 21:50

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .