没什么好说的，根据定义制剂代入，大到你无法想象的数

f(0,y)=y+1
f(1,y)=f(0,f(1,y-1)=f(0,f(0,..y层...f(1,0))=f(0,f(0,..y层...1)=y+2
f(2,y)=f(1,f(2,y-1)=f(1,f(1,..y层...f(2,0))=f(1,f(1,..y层...f(1,1))=f(1,f(1,..y层...f(2,0))=f(1,f(1,..y层...3)=2y+3
f(3,y)=f(2,f(2……y层..f(3,0))=f(2,f(2……y层..f(2,1)=f(2,f(2……y层.5)=2(2(...y层最内层为5..+3)))=2^(y+3)-3
继续第四层

其实，算几遍就很容易发现规律
f(1,y)=2+(y+3)-3
f(2,y)=2*(y+3)-3
f(3,y)=2^(y+3)-3
f(4,y)=2^2^2……y+3层……-3=2↑↑(y+3)-3
f(5,y)=2↑↑↑(y+3)-3
f(x,y)=2↑……x-2个↑(y+3)-3
所以F(4)=f(4,4)=2↑↑7-3