Hi Rakesh, > I think i made a mistake in presenting the function... The function > is > (x , x->y) -> z. > y is a function of x > z is a function of x and y This would have the MathType: ((x -> y), ((x, y) -> z)) However, it is odd that y occurs as both a dependent variable (in (x -> y)) and an independent variable (in ((x, y) -> z)). Cheers, Bill
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