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To make the proof more rigorous, we can start by defining the average number of hops between two random nodes in a network. Let H(N) be the average number of hops between two random nodes in a network with N nodes. Then we can define H(N) as follows:

$$H(N) = \frac{1}{N(N-1)} \sum_{i=1}^{N} \sum_{j=1, j \neq i}^{N} h(i,j)$$

where h(i,j) is the number of hops between nodes i and j.

Next, we can define a scale-free network as a network with a power-law degree distribution, where the number of nodes with degree k is given by:

$$P(k) \propto k^{-\gamma}$$

where P(k) is the probability of a node having degree k, and γ is a parameter that determines the shape of the distribution. For a scale-free network, γ is typically between 2 and 3.

Now, we can use these definitions to show that the average number of hops between two random nodes in a scale-free network is approximately equal to log(log(N)). To do this, we can make the following assumptions:

    The network is connected, meaning that there is a path between any two nodes.
    The number of hops between two nodes is the shortest path between them, i.e., the number of edges on the path.
    The degree of a node is independent of the degrees of other nodes, i.e., the probability of a node having degree k is not affected by the degrees of other nodes.

Under these assumptions, we can show that the average number of hops between two random nodes in a scale-free network is approximately equal to log(log(N)).

First, we can rewrite the definition of H(N) as follows:

$$H(N) = \frac{1}{N(N-1)} \sum_{i=1}^{N} \sum_{j=1, j \neq i}^{N} h(i,j)$$

$$= \frac{1}{N(N-1)} \sum_{i=1}^{N} \sum_{j=1, j \neq i}^{N} \min_{p \in P_{i,j}} \left| p \right|$$

where P_{i,j} is the set of paths between nodes i and j, and |p| is the number of edges on a path p.

Next, we can use the assumption that the degree of a node is independent of the degrees of other nodes to simplify this expression further:

$$H(N) = \frac{1}{N(N-1)} \sum_{i=1}^{N} \sum_{j=1, j \neq i}^{N} \min_{p \in P_{i,j}} \left| p \right|$$

$$= \frac{1}{N(N-1)} \sum_{i=1}^{N} \sum_{j=1, j \neq i}^{N} \min_{p \in P_{i,j}} \left( \sum_{e \in p} 1 \right)$$

$$=