Reduce[{x == z, y == z}] gives y == z && x == z but say I'm only interested in constraints that appear between x and y.

The desired result of this hypothetical reduce would be. x==y. Don't know if it helps, but I'm only working with equalities.

  • 3
    $\begingroup$ There's undocumented syntax for that specific case: Reduce[{x == z, y == z}, {x, y}, z]. $\endgroup$
    – J. M.'s torpor
    Apr 17 '20 at 3:33
  • $\begingroup$ Eliminate[{x == y, y == z}, z] $\endgroup$
    – wuyudi
    May 17 '20 at 4:18
  • $\begingroup$ @J.M. As with Solve, it's best practice to put the eliminated variable(s) in braces, to tell MMA that it's not a domain specification, e.g., Reduce[{x == z, y == z}, {x, y}, {z}]. $\endgroup$
    – theorist
    Jun 11 at 14:19
  • $\begingroup$ @theorist, yes, that's a recent development. $\endgroup$
    – J. M.'s torpor
    Jun 13 at 15:27

With Solve you can specify a list of variables to be eliminated. The list brackets are required even for a single variable to preclude interpretation as an attempt to specify a domain specification.

Solve[{x == z, y == z}, y, {z}][[1]]

(* {y -> x} *)

% /. Rule -> Equal

(* {y == x} *)
  • $\begingroup$ Is this documented? Where to find? $\endgroup$
    – Acus
    Jun 11 at 8:19
  • $\begingroup$ @Acus - It is an outdated capability that still works (presumably for backward compatibility). See Version 7 "Solve[eqns, vars, elims] attempts to solve the equations for vars, eliminating the variables elims." There are other uses of it on this forum. $\endgroup$
    – Bob Hanlon
    Jun 11 at 11:25

Eliminate seems to do the job, but dosent work well when the resulting constraint should be an inequality (for example y==x*x, eliminating x, I would like to get y>=0. The operation Resolve with Exists over the unwanted variables and Reals as domain should get the correct answer, the downside being that quantifier elimination might be computationally expensive.


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