Given S and K, I itself is unnecessary, since it can be built from the other two: ((S K K) x) = (S K K x) = (K x (K x)) = x for any term x.
给出S和K，I自身就不是必须的了，因为可以建造自其他两个： ((S K K) x) =(S K K x) = (K x (K x)) = x 对于任何的项x。