New Term example

From the cycling maths blog comes this problem:

Using the variables of average speed (A), speed uphill (su) and speed downhill (sd) the following expression can be formed

sd = A * su / (2*su - A) (1.0)

This is not a very revealing formula. Realising that speed uphill and downhill are fractions of the average speed a new expression can be formed where su is substituted for the expression A * p. Realising also in parallel that will give the opportunity to cancel the A. This then creates a much better formula expression the downhill speed as a fraction of the average speed and shows us the pure relationship between the uphill and downhill speeds.

sd = A * [ p / (2p - 1) ] (2.0)

Now the question is how to get a machine to simplify a problem by first making it more complicated! This is a very simple example but equation 2 was derived from equation 1 through this intermediate:

sd = A^2 * su / (2*A*su - A) (1.5)

It is a local maximum (if we could measure complexity as a continuous function) with the equation 2 as the minimum we seek and equation 1 being a higher minimum.

A tree search of all possible equations, Godel-style, would get there but would be extremely time consuming.

Doesn't the problem solver need to be semantic, and have a database so that it can understand what it is doing and trim the searches?

Anyway the example is logged.