help with substitution model [Sicp], using Clojure

I am studying the sicp book, and I have a doubt with the substitution model of a procedure:

(defn A
   [x,y]
     (cond (= y 0) 0
           (= x 0) (* 2 y)
           (= y 1) 2
           :else (A (- x 1) (A x (- y 1)))))

This procedure is part of the exercise 1.10. If I run the function in REPL with the following parameters (A 1 10), the result is 1024. I decided to verify the result using the Substitution Model, but the result was 2048.

This is the substitution model that I wrote. There is something wrong, but I don't know what.

(A 1 10)
(A (- 1 1) (A 1 (- 10 1))))
(A 0 (A 1 9)))
(A 0 (A (- 1 1) (A 1 (- 9 1)))))
(A 0 (A 0 (A 1 8))))
(A 0 (A 0 (A (- 1 1) (A 1 (- 8 1))))))
(A 0 (A 0 (A 0 (A 1 7)))))
(A 0 (A 0 (A 0 (A (- 1 1) (A 1 (- 7 1)))))))
(A 0 (A 0开发者_运维百科 (A 0 (A 0 (A 1 6)))))) 
(A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 6 1))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 5))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 5 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 4)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 4 1)))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 3))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (-3 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 2)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 2 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 2)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 2)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 4))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 4))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 8)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 8)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 16))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 16)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 32))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (* 2 32))))))
(A 0 (A 0 (A 0 (A 0 (A 0 64)))))
(A 0 (A 0 (A 0 (A 0 (* 2 64)))))
(A 0 (A 0 (A 0 (A 0 128))))
(A 0 (A 0 (A 0 (* 2 128))))
(A 0 (A 0 (A 0 256)))
(A 0 (A 0 (* 2 256)))
(A 0 (A 0 512))
(A 0 (* 2 512))
(A 0 1024)
2048 ????

Can anyone indicate what I did wrong? I am sorry for the length of the question.

Consider these lines:

(A 0 (A 0 (A 0 (A 1 7)))))
(A 0 (A 0 (A 0 (A (- 1 1) (A 1 (- 7 1)))))))
(A 0 (A 0 (A 0 (A 0 (A 1 6)))))) 
(A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 6 1))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 5))))))))

Strip off the redundant outer layers:

(A 1 7))
(A (- 1 1) (A 1 (- 7 1))))
(A 0 (A 1 6)))
(A 0 (A (-1 1) (A 1 (- 6 1)))))
(A 0 (A 0 (A 0 (A 1 5)))))

Somewhere in here you've ended up with mismatched parentheses, but that's not important. Note that in going from A 1 7 to A 1 6, a single outer layer of A 0 _ is created, as expected. In going from A 1 6 to A 1 5, you've got two new layers of A 0 _. Each of these ends up doubling the result, so that's why your answer is off by a factor of 2.