Inequality involving Abs[]

Question

Mathematica fails to do anything with

Reduce[Abs[x-3]<4, Reals]
(* Reduce[Abs[x-3]<4, Reals] *)

This answer contends that explicitly naming the variables to be reduced will solve the problem, and indeed

Reduce[Abs[x-3]<4, x, Reals]
(* -1 < x < 7 *)

But why is this the case? If either the absolute value is removed, or the < is changed to =, it is not necessary to provide the variable name.

Answer 1

After some back and forth with WRI, here is the answer: the domain restriction in Reduce[] applies to the specified variables and functions involving those variables. Since no variables are specified in my first example, the domain restriction effectively does nothing. That also explains why using {} does not solve the problem.

EDIT: After reading @Michael's comment below, I had more interactions with Wolfram support, and I am still somewhat confused. From WRI:

---

From Reduce documentation:

"Algebraic variables in expr free of the Subscript[x, i] and of each other are treated as independent parameters."

Since no Subscript[x, i] are specified, Reduce solves for the algebraic-level variables that are free of other algebraic-level variables.

In[1]:= Reduce`FreeVariables[x^2<4, "Algebraic"]
Out[1]= {x}

In[2]:= Reduce`FreeVariables[Abs[x-3]<4 && x\[Element]Reals, "Algebraic"]
Out[2]= {x}

In[3]:= Reduce`FreeVariables[Abs[x-3]<4, "Algebraic"]
Out[3]= {Abs[-3 + x]}

In[4]:= Reduce`FreeVariables[Re[x] > 0, "Algebraic"]
Out[4]= {Re[x]}

I guess the point is that x appears "algebraically" only in the first example; in the other examples it is the Abs or Re expression that is the "algebraic-level" variable.

This is still pretty unclear to me at a conceptual level (although I now know how to get the results I want), but I'm not sure what further questions to ask to clarify the situation. BTW, Reduce``FreeVariables is not documented.

Answer 2

I suspect Reduce treats the second argument as a variable. For instance:

Reduce[2 Reals == 1, Reals]
(* Reals == 1/2 *)

So I'm not sure there's anything wrong with

Reduce[Abs[x-3] < 4, Reals]
(* Reduce[Abs[x-3] < 4, Reals] *)

However, from the docs ("Details"):

Reduce[expr,vars,dom] restricts all variables and parameters to belong to the domain dom.
...
Algebraic variables in expr free of the Subscript[x, i] and of each other are treated as independent parameters.

I think this means that in Reduce[Abs[x-3]<4, {}, Reals], the x is not treated as a variable (to be solved for) but as a parameter, assumed to be real. Yet it still does not solve the inequality:

Reduce[Abs[x-3]<4, {}, Reals]
(* Abs[x-3]<4 *)

If you put the constraint in directly, you get the sought-after result:

Reduce[Abs[x - 3] < 4 && x \[Element] Reals, {}, Reals]
(* 1 < x < 7 *)

I would expect to get the same thing for Reduce[Abs[x-3]<4, {}, Reals], since x \[Element] Reals is supposed to be assumed.

It may be a bug and should be reported to WRI. See if they have an explanation.