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Question: Using a local variable breaks uneval

Question:Using a local variable breaks uneval

_Maxim_ 729
proc
November 21 2017
2 
proc() local a := 0, b := 1;
  print(proc(x::uneval) x := 1 end proc(a));
  print(proc(x::uneval) x := b end proc(a))
end proc();
                               1
Error, (in unknown) illegal use of a formal parameter

Why doesn't the second nested procedure keep a unevaluated?
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