How to distinguish x[a] from x in replacement rules? [closed]

Question

Quick to the point:is there a way to make Mathematica assume that $a \neq a[1] \neq a[2] , ...$ I notice that Solve assumes that $a$ is the same for all $a[1],a[2],...$ Also, if I replace $a \rightarrow b$ then I get all the $b[1],b[2],..$ for the whole expression.

Answer 1

Quoting from one of the comments above:

However, when you take the solution provided by Solve[{x + x[2] == 1}, {x}] and plug back to the equation you get something like (1- x[2] + (1-x[2])[2]), that is the main issue I'm having.

The easy fix is just changing x to something like x1 before using Solve but that might be a problem if the expressions are lengthy . Here are some replacement rules that change x to x1 while keeping x[2]:

Reference:

Expand[(x + Cos[x[2] + x]*x[2]*x + Sin[x[3]*x])^2 + x]

x+x^2+2 x Sin[x x[3]]+Sin[x x[3]]^2+2 x^2 Cos[x+x[2]] x[2]+2 x Cos[x+x[2]] Sin[x x[3]] x[2]+x^2 Cos[x+x[2]]^2 x[2]^2

• Option 1

Straightforward but clunky as it introduces an auxiliary variable:

 expression1=
 (Expand[(x + Cos[x[2] + x]*x[2]*x + Sin[x[3]*x])^2 + x]
 /. x[s_] :> b[s] /. x :> x1 /. b :> x)

x1+x1^2+2 x1 Sin[x1 x[3]]+Sin[x1 x[3]]^2+2 x1^2 Cos[x1+x[2]] x[2]+2 x1 Cos[x1+x[2]] Sin[x1 x[3]] x[2]+x1^2 Cos[x1+x[2]]^2 x[2]^2

• Option 2

Quick but mysterious :

expression2=
(Expand[(x + Cos[x[2] + x]*x[2]*x + Sin[x[3]*x])^2 + x] 
/. {x -> x1, x[s_] :> x[s]});

It's not entirely clear to me why that works. I think it's because ReplaceAll works from the outside in as explained here https://mathematica.stackexchange.com/a/188608/86543 but I feel unsafe using that as I am not entirely sure.

• Option 3

Convenient:

(The code below uses the resource function ReplaceAllOutside)

 expression3=
(Expand[(x + Cos[x[2] + x]*x[2]*x + Sin[x[3]*x])^2 + x] // 
 ResourceFunction["ReplaceAllOutside"][x :> x1, x[_]]);

Check:

expression1 == expression2 == expression3

True