How to avoid Replace substituting subscripts?

Question

I have a polynomial in $\rho$ with many parameters (I will omit the polynomial code in the Mathematica expressions below because it is too long):

$$ \rho ^3 J (\mu +1) \left[(A \delta \tau _0+B (1-\delta ) \tau _1\right]+ \rho ^2 \left[\left(A \delta \tau _0+B (1-\delta ) \tau _1\right) \left((\mu +1) V_B-J\right)+A (1-\delta ) J (\mu +1) \tau _0+B \delta J (\mu +1) \tau _1+\delta \mu ^2 \tau _0+(1-\delta ) \mu ^2 \tau _1\right]+ \rho \left[A (1-\delta ) \tau _0 \left((\mu +1) V_B-J\right)-V_B \left(A \delta \tau _0+B (1-\delta ) \tau _1\right)+B \delta \tau _1 \left((\mu +1) V_B-J\right)+(1-\delta ) \mu ^2 \tau _0+\delta \mu ^2 \tau _1\right] -A (1-\delta ) \tau _0 V_B-B \delta \tau _1 V_B $$

Notice that $B$ is both a parameter and an index for $V_B$.

Say I saved this polynomial in p, then when I use Replace using levelspec, such as:

Replace[p, {δ -> 0, B -> 0}, 6]

Mathematica substitutes both $B$ in the coefficients and in the subscripts (why would anyone want this?):

$$ \rho \left(A \tau _0 \left((\mu +1) V_B-J\right)+\mu ^2 \tau _0\right)+\rho ^2 \left(A J (\mu +1) \tau _0+\mu ^2 \tau _1\right)-A \tau _0 V_0 $$

Notice now there's a $V_0$ and a $V_B$. If I don't use levelspec, and use /. instead, all $B$ are replaced, in all indices (which is obvisouly what I do not want).

How to avoid subscripts substitution?

polynomial code is here:

p=-A (1 - δ) Subscript[V, B] Subscript[τ, 0] - B δ Subscript[V, B] Subscript[τ, 1] + J (1 + μ) ρ^3 (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]) + ρ ((1 - δ) μ^2 Subscript[τ, 0] + A (1 - δ) (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 0] + δ μ^2 Subscript[τ, 1] + B δ (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 1] - Subscript[V, B] (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ,1])) + ρ^2 (δ μ^2 Subscript[τ, 0] + A J (1 - δ) (1 + μ) Subscript[τ, 0] + (1 - δ) μ^2 Subscript[τ, 1] + B J δ (1 + μ) Subscript[τ, 1] + (-J + (1 + μ) Subscript[V, B]) (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]))

Answer 1

The real solution here is not to use B in a subscript, or better: not to use Subscript at all.

If you can't or don't want to do that, you can add an extra transformation rule for all Subscript expressions. Once ReplaceAll processes a subexpression, it won't touch it anymore.

expr = Subscript[x, B] + B;

expr /. B -> 1
(* 1 + Subscript[x, 1] *)

expr /. {s_Subscript :> s, B -> 1}
(* 1 + Subscript[x, B] *)

Answer 2

Mathematica does not concern itself with why you might want to do something, it does what you indicate that you want done.

Convert the subscripts to strings prior to replacing B

p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]], δ -> 0, 
  B -> 0}

Or, if you don't want the subscripts left as strings

p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]], 
  δ -> 0, B -> 0} /. 
    {Subscript[var_, sub__] :> Subscript[var, ToExpression[sub]]}

Answer 3

Try this:

    p /. Subscript[a_, B] -> Subscript[a, x] /. {δ -> 0, 
   B -> 0} /. Subscript[a_, x] -> Subscript[a, B]

yielding the following

Have fun!