# [tlaplus] Re: Removing "equivalents" from a set

Thank you for the reply! I actually didn't think of SUBSET because I never really thought of using it for anything else than a set of numbers! I tried it on a very small set derived from 1..3, but I haven't got a result in a couple of minutes. I have decided that the of couple extra elements will not hurt my model as much as this heavy computation. I still appreciate the answer, I'm sure it will help me down the road. Also I think you missed that my resulting set contains element {{2}, {3, 4}}  which has the union {2, 3, 4}.

On Friday, April 19, 2019 at 12:01:51 AM UTC+2, Leslie Lamport wrote:
In constructing the second set, you have removed from s the element {{3}, {2,4}}  the union of the elements of that set is the set {2,3,4}.  That set is not in s, since s has no 3-element sets, so it does not satisfy your condition for an element to be removed from s.  Therefore, what you have written makes no sense to me.

You seem to want to define an operator Op such that Op(s) equals the second set.  By talking about "removing elements", I presume you meant that Op(s) should be a subset of s.  Hence, I believe that the definition you want should be expressible as

Op(t) == CHOOSE u \in SUBSET t : P(t, u)

for some formula  P(t, u).  This definition of Op may not satisfy you because to evaluate it, TLC must compute all elements of SUBSET t and evaluate P(t,u) for all the elements u of SUBSET t until it finds one that evaluates to TRUE.  However, it will at least tell people what operator Op you want.

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