To solve the optimization problem described, we first need to translate the natural language description into a symbolic representation. This involves defining variables and formulating the objective function and constraints using these variables.

Let's define:
- \(x_1\) as the number of regular models produced,
- \(x_2\) as the number of premium models produced.

The objective is to maximize profit. Given that each regular model yields a profit of $200 and each premium model yields a profit of $500, the objective function can be written as:
\[ \text{Maximize:} \quad 200x_1 + 500x_2 \]

The constraints are based on the available time for assembly and software verification. For assembly, the total time used by regular models (\(70x_1\)) and premium models (\(100x_2\)) must not exceed 6000 minutes:
\[ 70x_1 + 100x_2 \leq 6000 \]

For software verification, the total time used by regular models (\(25x_1\)) and premium models (\(30x_2\)) must not exceed 4000 minutes:
\[ 25x_1 + 30x_2 \leq 4000 \]

Additionally, \(x_1\) and \(x_2\) must be non-negative since they represent the number of units produced.

Thus, the symbolic representation in JSON format is:
```json
{
    'sym_variables': [('x1', 'number of regular models'), ('x2', 'number of premium models')],
    'objective_function': '200*x1 + 500*x2',
    'constraints': ['70*x1 + 100*x2 <= 6000', '25*x1 + 30*x2 <= 4000', 'x1 >= 0', 'x2 >= 0']
}
```

Now, let's write the Gurobi code to solve this optimization problem:
```python
from gurobipy import *

# Create a new model
m = Model("Theta_Electronics")

# Define variables
x1 = m.addVar(name="regular_models", vtype=GRB.INTEGER, lb=0)
x2 = m.addVar(name="premium_models", vtype=GRB.INTEGER, lb=0)

# Set the objective function
m.setObjective(200*x1 + 500*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(70*x1 + 100*x2 <= 6000, name="assembly_time")
m.addConstr(25*x1 + 30*x2 <= 4000, name="software_verification_time")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print(f"Optimal solution found: {x1.varName} = {x1.x}, {x2.varName} = {x2.x}")
    print(f"Maximum profit: {m.objVal}")
else:
    print("No optimal solution found")
```