Dyscopes are currently stored in an array in the state.
This gives each an ID.
Reachable dyscopes: call stacks always have reference to concrete dyscope.
Pointers to anywhere:
equality of two dyscopes:
Type of value associated to a dyscope: struct {int}
d1==d2  becomes d1.0==d2.0.   This one survives renumbering, but not
d1<d2.  Sure, that becomes an uninterpreted function.
What about

d1==d2 --> d1.0==5.

Renumbering: Apply permutation to d1.0?


f(2)=f(3) ... need that to return false

CIVL:evaluation of equality for dyscope types special handling
 (also inequality, <, >)


A concrete state has some concrete ordered list of dyscopes
(0,1,2,...,n-1)
and values of expressions of dyscope type are integers.
Two concrete states are equivalent if there is a permutation
of the integers 0..n-1 such that when applied to one state
by permuting the dyscopes in the list and the id numbers in the values,
you get the other state.

A symbolic state represents a set of concrete states.
Question is given two symbolic states, are they "equivalent"
meaning given any concrete state cs1 in ss1, there is a concrete
state cs2 in ss2 such that cs1 and cs2 are equivalent.
And vice versa.

what is the permutation applied to the symbolic state?

var: dyscope ... 

var: array of T..

example: (0 1 2)

SS1:
d1: X
d2: Y
d3: <1>
d4: Z
pc: X<Y && Y<1 && Z=2

To show: let f be the permutation 
 X<Y iff f(X)<f(Y)
 X=Y iff f(X)=f(Y)

SS2:
d1: X
d2: Y
d3: <2>
d4: Z
pc: X<Y && Y<2 && Z=0


evaluation of operations on pointers or anything involving

f(3)=X
f(3)=f(4)

X: arbitrary pointer
X.0 : dyscope type
X.0==f(3)

Can we guarantee that f will only be applied to concrete integers?

"I never want to see a conrete integer representing a dyscope taken out of f"