Clarify that bounded comparison evaluates x only once

The old definition

    l<x<u means x>l && x<u

could have been misunderstood to mean that x would be evaluated twice. Rectify this using 'let'.

@gavinking do you think this is necessary? Not sure.

(It also seems that the compiler implements `l<x<u` as `l<x && x<u`, rather than `x>l && x<u`, but I’m starting to feel like an obnoxious nit-pick pointing out these small differences, so I’m just going to forget that I saw that. No need to delay 1.1 for that.)