To solve the given optimization problem using Gurobi, we first need to understand and translate the natural language description into a mathematical formulation. The objective is to maximize the function: $7 \times \text{cornichons} + 1 \times \text{strips of bacon} + 1 \times \text{slices of pizza}$.

Given constraints:
- Each cornichon contains 7 grams of fiber.
- Each strip of bacon contains 6 grams of fiber.
- Each slice of pizza contains 1 gram of fiber.
- The total fiber from cornichons and strips of bacon must not exceed 22 grams.
- The total fiber from cornichons and slices of pizza must not exceed 22 grams.
- The total fiber from all three (cornichons, strips of bacon, and slices of pizza) must not exceed 23 grams.

Let's denote:
- $x_0$ as the number of cornichons,
- $x_1$ as the number of strips of bacon,
- $x_2$ as the number of slices of pizza.

The constraints can be formulated as follows:
1. $7x_0 + 6x_1 \leq 22$
2. $7x_0 + x_2 \leq 22$
3. $7x_0 + 6x_1 + x_2 \leq 23$

Since $x_0$ (cornichons) and $x_1$ (strips of bacon) can be non-integer, but $x_2$ (slices of pizza) must be an integer, we will use continuous variables for $x_0$ and $x_1$, and an integer variable for $x_2$.

Here is the Gurobi code to solve this optimization problem:

```python
from gurobipy import *

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

# Define variables
x0 = m.addVar(lb=0, ub=GRB.INFINITY, vtype=GRB.CONTINUOUS, name="cornichons")
x1 = m.addVar(lb=0, ub=GRB.INFINITY, vtype=GRB.CONTINUOUS, name="strips_of_bacon")
x2 = m.addVar(lb=0, ub=GRB.INFINITY, vtype=GRB.INTEGER, name="slices_of_pizza")

# Define constraints
m.addConstr(7*x0 + 6*x1 <= 22, name="fiber_from_cornichons_and_bacon")
m.addConstr(7*x0 + x2 <= 22, name="fiber_from_cornichons_and_pizza")
m.addConstr(7*x0 + 6*x1 + x2 <= 23, name="total_fiber")

# Define the objective function
m.setObjective(7*x0 + x1 + x2, GRB.MAXIMIZE)

# Optimize the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"cornichons: {x0.x}")
    print(f"strips_of_bacon: {x1.x}")
    print(f"slices_of_pizza: {x2.x}")
    print(f"Objective value: {m.objVal}")
else:
    print("No optimal solution found")
```