python-mip: Incremental Solving not supported?

Hello, I’ve been trying to learn how to do boolean modelling in Mixed Integer Programming, and I would like to get all solutions to the problem. Here is a simple example model where blocks with ports need to be connected, with some symmetry breaking on the blocks and ports (should have 3 solutions):

import mip

mm = mip.Model(solver_name="cbc")
mm.verbose = False

# Blocks
bus = mm.add_var("bus", 0, 1, var_type=mip.BINARY)
battery_1 = mm.add_var("battery_1", 0, 1, var_type=mip.BINARY)
battery_2 = mm.add_var("battery_2", 0, 1, var_type=mip.BINARY)

# Ports
bus_p1 = mm.add_var("bus_p1", 0, 1, var_type=mip.BINARY)
bus_p2 = mm.add_var("bus_p2", 0, 1, var_type=mip.BINARY)
battery_1_p = mm.add_var("battery_1_p", 0, 1, var_type=mip.BINARY)
battery_2_p = mm.add_var("battery_2_p", 0, 1, var_type=mip.BINARY)

# Constraints
mm.add_constr(battery_1 + battery_2 >= 0)
mm.add_constr(battery_1 + battery_2 <= 2)
mm.add_constr(bus == 1)
mm.add_constr(battery_1 >= battery_2)
mm.add_constr(battery_1 == battery_1_p)
mm.add_constr(battery_2 == battery_2_p)
mm.add_constr(bus >= bus_p1)
mm.add_constr(bus >= bus_p2)
mm.add_constr(bus_p1 <= bus_p2)
mm.add_constr(bus_p1 + bus_p2 == battery_1_p + battery_2_p)

count = 0
for i in range(0, 10):
    status = mm.optimize()
    if status == status.FEASIBLE or status == status.OPTIMAL:
        print(f"{bus}: {bus.x}")
        print(f"{battery_1}: {battery_1.x}")
        print(f"{battery_2}: {battery_2.x}")
    else:
        print("INFEASIBLE")
        break
    count += 1
    # or(a != a.x, b != b.x, c != by)
    flip = [1 - v if v.x else v for v in (bus, battery_1, battery_2)]
    mm.add_constr(mip.xsum(flip) >= 1)
print(f"CBC (Python-MIP) found {count} solutions")

However this code only gives me 1 solution, 10 times.

If I do the same with google Or-Tools, it works as expected:

from ortools.linear_solver import pywraplp as or_tools

om = or_tools.Solver("cbc", problem_type=or_tools.Solver.CBC_MIXED_INTEGER_PROGRAMMING)

# Instances
bus = om.BoolVar("bus")
battery_1 = om.BoolVar("battery_1")
battery_2 = om.BoolVar("battery_2")

# Ports
bus_p1 = om.BoolVar("bus_p1")
bus_p2 = om.BoolVar("bus_p2")
battery_1_p = om.BoolVar("battery_1_p")
battery_2_p = om.BoolVar("battery_2_p")

# Constraints
om.Add(battery_1 + battery_2 >= 0)
om.Add(battery_1 + battery_2 <= 2)
om.Add(bus == 1)
om.Add(battery_1 >= battery_2)
om.Add(battery_1 == battery_1_p)
om.Add(battery_2 == battery_2_p)
om.Add(bus >= bus_p1)
om.Add(bus >= bus_p2)
om.Add(bus_p1 <= bus_p2)
om.Add(bus_p1 + bus_p2 == battery_1_p + battery_2_p)

count = 0
# Limit number of solutions
for i in range(0, 10):
    status = om.Solve()
    if status == om.FEASIBLE or status == om.OPTIMAL:
        print(f"{bus}: {bus.solution_value()}")
        print(f"{battery_1}: {battery_1.solution_value()}")
        print(f"{battery_2}: {battery_2.solution_value()}")
    else:
        print("INFEASIBLE")
        break
    count += 1
    # or(a != a.x, b != b.x, c != by)
    flip = [1 - v if v.solution_value() else v for v in (bus, battery_1, battery_2)]
    om.Add(sum(flip) >= 1)
print(f"CBC (Or-Tools) found {count} solutions")

This gives the expected 3 solutions.

Is there anything I’m missing, like a different way to restrict solutions, or is python-mip just not intended to be used like this?