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Here we have another conflict between a free variable and one which is lambda-bound.
The conflict exists between the `x' in `(lambda(x)'
and the `x' in the expression `(x(x(x x0)))'. To dissolve the
conflict the lambda-bound `x' has been renamed `x1'.
Synthesis of (lambda(f) ...) and `(lambda(x0) ...)' in 
Now the synthesis can be performed:
Georg P. Loczewski