ReplaceAll works and does not work for the same expression

Question

I will give two cases, where the rule replacement works in the first one and does not work in the second one, but they are essentially the same!
Consider the following definition, whose details are not important:
vall = -Sqrt[Δ^2 + μ^2 + 2*μ*Sqrt[px^2 + py^2] + px^2 + py^2]

Now, we check the following:
vall /. {ϕ -> 0, py -> pyy, px -> 0, μ -> 0}

As expected, I can obtain the desired replacement:

Now, consider the following expression, which contains the above expression as a subexpression:
Exists[
 {py, pyy}, 
 123 == (vall /. {ϕ -> 0, py -> pyy, px -> 0, μ -> 0}) == 0  ]

The result is surprising:

As can be seen in the picture, no replacement py -> pyy is made!
Why does this happen?

MarcoB · Accepted Answer

I suspect that this has to do with the HoldAll attribute of Exists (see Attributes[Exists]). However, if you force evaluation of its second argument, i.e. the expression you are trying to modify, you obtain the desired result:
Exists[
  {py, pyy}, 
  Evaluate[ 123 == (vall /. {ϕ -> 0, py -> pyy, px -> 0, μ -> 0}) == 0 ]
]

(* Out: Exists[{pyy}, 123 == -Sqrt[pyy^2 + Δ^2] == 0] *)

ReplaceAll works and does not work for the same expression

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