I'll show you several ways to solve this problem.

First Version: List Comprehensions

check_products(AmountOfProducts, OrderProducts) ->
  [] ==
    [Key || {Key, _} <- OrderProducts
          , not lists:keymember(Key, 1, AmountOfProducts)].

Here we basically collect all the keys in OrderProducts that don't have a corresponding tuple in AmountOfProducts and, if that's an empty list… we return true

Second Version: High-Order Functions

check_products(AmountOfProducts, OrderProducts) ->
  lists:all(
    fun({Key, _}) ->
      lists:keymember(Key, 1, AmountOfProducts)
    end, OrderProducts).

Here we basically write what you wrote in your question, but in Erlang code: Return true if the keys of all the elements in OrderProduct are also keys in AmountOfProducts .

Third Version: Just compare the Keys

Now, if you know that there will be no duplication of keys in any of the lists, you can also write your code like this…

check_products(AmountOfProducts, OrderProducts) ->
  [] == keys(OrderProducts) -- keys(AmountOfProducts).

Basically saying that if we remove all the keys from AmountOfProducts from the list of keys of OrderProducts ... we end up with no keys (they were all removed because they belonged to AmountOfProducts ). You have to implement an auxiliary function keys/1 , but that one is hopefully much easier to implement (or you can just use proplists:get_keys/1.