7
$\begingroup$

This is quite surprising to me.

Simplify[a == b, {a == {1}, b == {1}}]
a == b

does not evaluate the equality, while

Simplify[a == b, {a == 1, b == 1}]
True

does. What is the explanation?

Another one:

Simplify[a/b == e/f, {a/b == c/d, c/d == e/f}]
a/b == e/f
$\endgroup$
  • 2
    $\begingroup$ Even FullSimplify can't do this. But I found this: In[] Simplify[a == b, {a == c[{1}], b == c[{1}]}] Out[] True $\endgroup$ – wacharin wichiramala Mar 15 '16 at 8:53
  • 3
    $\begingroup$ Perhaps an even more succinct example is comparing Simplify[a, a == {1}] to Simplify[a, a == 1] $\endgroup$ – Jason B. Mar 15 '16 at 9:16
  • $\begingroup$ List equality/simplification vs number equality/simplification, interesting. $\endgroup$ – barrycarter Mar 15 '16 at 15:13
  • 1
    $\begingroup$ But isn't this a matter of {1} not being "more simple" than a? Simplify does what it says: it simplifies an expression according to some "aesthetic". One thing that it does is to minimize the LeafCount of the expression, and LeafCount@a is 1 whereas LeafCount@{1} is 2. @JasonB. $\endgroup$ – march Mar 15 '16 at 16:52
  • $\begingroup$ @JasonB and OP: I recommend putting those examples into the original post, because especially the OP's version with c[{1}] is confusing (although I think my previous comment accounts for JasonB's example and it might account for the OP's). $\endgroup$ – march Mar 15 '16 at 16:58

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