proof

How many binary search trees can be constructed from n distinct elements? And how can we find a mathematically proved formula for it? Example: If we have 3 distinct elements, say 1, 2, 3, there are 5 binary search trees. Given n elements, the number of binary search trees that can be made from those elements is given by the nth Catalan number (denoted C n ). This is equal to Intuitively, the Catalan numbers represent the number of ways that you can create a structure out of n elements that is made in the following way: Order the elements as 1, 2, 3, ..., n. Pick one of those elements to use as

I'm now learning Kademlia network by reading the classical paper Kademlia: A Peer-to-peer Information System Based on the XOR Metric . I want to understand the complexity of its operation but still cannot figure it out. In the 3 Sketch of proof section, the paper gives two definitions: Depth of a node (h) : 160 − i, where i is the smallest index of a non-empty bucket Node y’s bucket height in node x : the index of the bucket into which x would insert y minus the index of x’s least significant empty bucket . And three conclusions: With overwhelming probability the height of a any given node

I want to prove or falsify forall (P Q : Prop), (P -> Q) -> (Q -> P) -> P = Q. in Coq. Here is my approach. Inductive True2 : Prop := | One : True2 | Two : True2. Lemma True_has_one : forall (t0 t1 : True), t0 = t1. Proof. intros. destruct t0. destruct t1. reflexivity. Qed. Lemma not_True2_has_one : (forall (t0 t1 : True2), t0 = t1) -> False. Proof. intros. specialize (H One Two). inversion H. But, inversion H does nothing. I think maybe it's because the coq's proof independence (I'm not a native English speaker, and I don't know the exact words, please forgive my ignorance), and coq makes it