$($\%l x.x) 1 = x[x:=1] = 1
  $($\%l x.x+1) 2 = (x+1)[x:=2] = 2 + 1
  $($\%l x.x+y+1) 3 = (x+y+1)[x:=3] = 3+y+1
  $($\%l y.(\%l x.x+y+1) 3) 4) = 
  		$(($\%l x.x+y+1) 3)[y:=4] = 3 + 4 + 1