It might help to simplify. Whenever you're accumulating over a
sequence of things, think of reduce:
(let [__ (fn [& fs]
;; Here's the function:
(reduce #(fn [x] (%1 (%2 x))) fs))
]
;; Testing:
[ (= 5 ((__ (partial + 3) second) [1 2 3 4]))
(= [3 2 1] ((__ rest reverse) [1 2 3 4]))
])
Each step in the accumulation creates a new function that just applies
the current function to an argument that is the result of applying the
already-composed ones.
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