Foldl in Haskell to SKI combinator

[TO]

Hi,

Please help me with your answer.
Problem:
Convert the lambda function below to SKI combinators.
\xyz->x(xyz)
(foldl in Haskell)

Thanks,

TO

[TO]

Post by TO
Hi,
Please help me with your answer.
Convert the lambda function below  to SKI combinators.
   \xyz->x(xyz)
(foldl in Haskell)
Thanks,
TO

I'm sorry but I've foggeten Curring is needed for x. Foldl is
converted from B=S(KS)K

Thanks,

TO