For python there is a NATO dictionary already.
Can I suggest you print it out in the demo program, it'll save people typing out the whole dictionary.

The description is not correct.
they have fib(0) = 1
and there code doesn't calculate the fibonnaci series.

Here is my alternative expansion.

The function is recursively defined by:

0.    fib(0) = 0
1.    fib(1) = 1
2.    fib(2) = 1
3.    fib(n + 2) = fib(n) + fib(n + 1), if n + 2 > 1

3.    fib(n + 1) + fib(n + 2) = fib(n) + 2 * fib(n + 1)

The two sides of the equation look much more alike, but there is still an essential difference, which is the coefficient of the second term of each side. On the left side of the equation, it is 1 and, on the right, it is 2. To remedy this, we can introduce a variable b:

3.     b * fib(n + 1) + b * fib(n + 2) = b * fib(n) + (b + b) * fib(n + 1)

We then subtract both sides by b * fib(n+1)

3.     b * fib(n + 2) = b * fib(n) + (b + b)

To try and return the tail to the function we add both sides by a * fib(n+1) to each side:

3.     a * fib(n + 1) + b * fib(n + 2) = b * fib(n) + (a + b) * fib(n + 1)

Now the two sides have the same form (call it F), this function has the form:

3.     F(c, d, e) = F(a, b, n+1) = F(a, a+b, n) = c * fib(e) + d * fib(e + 1)

(Note if your confused by the above equation substitute the values into the furthest right function and you get)

3.     a * fib(n + 1) + b * fib(n + 2) = b * fib(n) + (a + b) * fib(n + 1) = c * fib(n) + d * fib(n + 1)

Which we can simplify to:

3.     F(a, b, n + 1) = F(b, a + b, n)

We also have, by definition of F and fib:

4.     F(a, b, 0) = a * fib(0) + b * fib(1) = b

(Proove this with general equation)
F(a, b, n + 1) with n = 0
c * fib(n) + d * fib(n + 1)
a * fib(0) + b * fib(0 + 1) as fib(0) = 0, fib(1) = 1
b

(Proove this with general equation)
c * fib(e) + d * fib(e + 1)
1 * fib(n) + 0 * fib(n + 1)
1 * fib(n)

(Proove this with general equation)
F(a, b, n + 1) with n = 1
c * fib(n) + d * fib(n + 1)
a * fib(1) + b * fib(1 + 1) as fib(1) = fib(2) = 1
a + b

def fib(n):
    if n == 0: return 0    # see 6. above
    def F(a, b, n):
        if n == 1: return a + b    # see 6. above
        return F(b, a + b, n - 1)  # see 3. above (F(a, b, n + 1) = F(b, a + b, n)) use n = n - 1
    return F(1, 0, n)              # see 5. above

The final step (optional for languages that support tail call optimization) is to replace the tail recursive function F with a loop:

def fib(n):
    if n == 0: return 0            # see 4. above
    a, b = 1, 0                    # see 5. above
    while n > 1:		
        a, b, n = b, a + b, n - 1  # see 3. above
    # n = 1 at this point
    return a + b<>                   # see 6. above 

Your fibonacci description is very wrong, and the code at the end doesn't work:
A big flaw in your description starts here:
1. fib(0) = 1

I did the algebra in a bit more detail trying to work out what went wrong.
So here is my description, based on what you had:

The function is recursively defined by:

0.    fib(0) = 0
1.    fib(1) = 1
2.    fib(2) = 1
3.    fib(n + 2) = fib(n) + fib(n + 1), if n + 2 > 1

3.    fib(n + 1) + fib(n + 2) = fib(n) + 2 * fib(n + 1)

The two sides of the equation look much more alike, but there is still an essential difference, which is the coefficient of the second term of each side. On the left side of the equation, it is 1 and, on the right, it is 2. To remedy this, we can introduce a variable b:

3.     b * fib(n + 1) + b * fib(n + 2) = b * fib(n) + (b + b) * fib(n + 1)

We then subtract both sides by b * fib(n+1)

3.     b * fib(n + 2) = b * fib(n) + (b + b)

To try and return the tail to the function we add both sides by a * fib(n+1) to each side:

3.     a * fib(n + 1) + b * fib(n + 2) = b * fib(n) + (a + b) * fib(n + 1)

Now the two sides have the same form (call it F), this function has the form:

3.     F(c, d, e) = F(a, b, n+1) = F(a, a+b, n) = c * fib(e) + d * fib(e + 1)

(Note if your confused by the above equation substitute the values into the furthest right function and you get)

3.     a * fib(n + 1) + b * fib(n + 2) = b * fib(n) + (a + b) * fib(n + 1) = c * fib(n) + d * fib(n + 1)

Which we can simplify to:

3.     F(a, b, n + 1) = F(b, a + b, n)

We also have, by definition of F and fib:

4.     F(a, b, 0) = a * fib(0) + b * fib(1) = b

(Proove this with general equation)
F(a, b, n + 1) with n = 0
c * fib(n) + d * fib(n + 1)
a * fib(0) + b * fib(0 + 1) as fib(0) = 0, fib(1) = 1
b

(Proove this with general equation)
c * fib(e) + d * fib(e + 1)
1 * fib(n) + 0 * fib(n + 1)
1 * fib(n)

(Proove this with general equation)
F(a, b, n + 1) with n = 1
c * fib(n) + d * fib(n + 1)
a * fib(1) + b * fib(1 + 1) as fib(1) = fib(2) = 1
a + b

def fib(n):
    if n == 0: return 0    # see 6. above
    def F(a, b, n):
        if n == 1: return a + b    # see 6. above
        return F(b, a + b, n - 1)  # see 3. above (F(a, b, n + 1) = F(b, a + b, n)) use n = n - 1
    return F(1, 0, n)              # see 5. above

The final step (optional for languages that support tail call optimization) is to replace the tail recursive function F with a loop:

def fib(n):
    if n == 0: return 0            # see 4. above
    a, b = 1, 0                    # see 5. above
    while n > 1:		
        a, b, n = b, a + b, n - 1  # see 3. above
    # n = 1 at this point
    return a + b<>                   # see 6. above