Is possible in gremlin to count something inside a choose and then go back to a previous step? - gremlin

Question

Is possible in gremlin to count something inside choose() and then go back to a previous step in the traversal?:
g.V()
.hasLabel("user1")
.as("firstStep")
.V()
.as("secondStep")
.fold()
.choose(
count(local).is(gt(1)),
select('firstStep'),
select('secondStep')
)
The select('firstStep') line is executed but nothing is returned because at that point everything stored by .as("firstStep") was removed by fold()
Of course I can always solve this kind of problem at the expense of performance, duplicating the cost by searching 2 times:
g.V()
.hasLabel("user1")
.choose(
V().fold().count(local).is(lt(2)),
V().fold()
)
I'm looking for a better solution than this.
This is like if I go to a store and pay something, because I have to count the change I forget how to go back to my hose, the gremlins of the movie are more intelligent than that. I hope there is a solution that I ignore.
EDIT:
There is an answer recommending to replace as() by store() but that is not a general solution, doesn't work in most situations becase these are very different tools, for example this query has the same problem and replacing as() by store() gives a different result:
g.V()
.hasLabel('user')
.project("user")
.by(
as("singleUser")
.V()
.fold()
.choose(
count(local).is(gt(1)),
select('singleUser'),
unfold()
)
)

Answer 1

In this pattern you can use store step, that way the "firstStep" will be saved after the fold.
g.V().hasLabel('user1').store('firstStep').
V().
fold().choose(
count(local).is(gt(1)),
select('firstStep').unfold(),
select('secondStep')
)
example: https://gremlify.com/mg7noyf1cdq