I want to show the following equality (using logical connectives, not venn diagrams)

Show that: $$(A ∩ B) ▵ C = (A ▵ C) ▵ (A \setminus B)$$

$A ▵ B$ is defined as: $(A ∪ B) \setminus (A ∩ B)$

My attempt:

If I expand the symmetric difference symbol $▵$ in the original equality I get:

$$((A ∩ B) ∪ C) \setminus ((A ∩ B) ∩ C) =$$

$$(((A ∪ C) \setminus (A ∩ C)) ∪ (A \setminus B)) \setminus (((A ∪ C) \setminus (A ∩ C)) ∩ (A \setminus B))$$

So now I want to write the second statement in the equality with logical connectives and reduce it to the first one:

$$((A \land B) \lor C) \land \lnot ((A \land B) \land C) =$$

$$(((A \lor C) \land \lnot (A \land C)) \lor (A \land \lnot B)) \land \lnot (((A \lor C) \land \lnot (A \land C)) \land (A \land \lnot B)) =$$

But I am stuck here. I don't really know where to begin...

  • $\begingroup$ It would seem that you swapped the intersection with the union on the LHS of the equation when you expanded the symmetric difference. Also, changing to logical notation is unnecessary. Just note that $x\setminus y=x\cap y^C$ and use DeMorgan's Laws (set version) for the complements. If you insist on using the logical form, DeMorgan's laws are still going to come into play. $\endgroup$
    – Terra Hyde
    Jul 22, 2015 at 19:23
  • $\begingroup$ Thank you for your comment @TerraHyde , if you have time I would love to see how you would do the development using set operations. $\endgroup$ Jul 22, 2015 at 20:06
  • $\begingroup$ Where did you encounter this horrible thing? $\endgroup$ Jul 22, 2015 at 20:54
  • $\begingroup$ Hi Daniel. It is from the book "How to prove it". The question said that I could prove the equality by any method, Venn diagrams was a bit easy so I thought about using logical connectives instead. $\endgroup$ Jul 22, 2015 at 21:59
  • $\begingroup$ I will work on the full solution using set operations, but it may take a while. I have done this problem before, but I forgot the particulars for the middle steps. $\endgroup$
    – Terra Hyde
    Jul 23, 2015 at 3:33

1 Answer 1


When things are finite and you are in doubt, resort to brute-force (though, this isn't actually good advice...)

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  • $\begingroup$ Hahaha thank you! Not exactly the kind of answer I was looking for but this is a good method to know about. $\endgroup$ Jul 22, 2015 at 19:44

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