Proof for the algorithm for m=n case. Let name middle element as "k"
 m1:=median (A, N)    -   O (1)
 m2:=median (B, N)    -   O (1)
Proof:
If m1 < m2, so m1 < k < m2 otherwise m2 < k < m1
Without loss of generality –if  m1 < k < m2, we know that the median of the two arrays must be
Must be A[m1] < k <B [m2] so we can drop the irrelevant parts (A[0….m1] , B[m2……N]) by searching for new medians in each array.

Proof by contradiction:
    Again without loss of generality…
        We will assume that k is found within ((A[0….m1] U B[m2……N]))
        Array A contains (((N+1)/2) -1) elements smaller than m1
        Array B contains maximum (((N+1)/2) -1) elements smaller than m2
        Unified set is -  ((A[0….m1] U B[m2……N]))   :
        (((N+1)/2) -1) + (((N+1)/2) -1) = N-1   elements before m1
        Contradicting the definition of median which is N= [((2N+1) / 2)]
        We can come to a conclusion that k cannot be a part of  ((A[0….m1] U B[m2……N]))
        Therefore it must be found somewhere  ((A[m1….n] U B[0….m2]))

m1:=median (A+m1, N / 2)    -   O (1)
m2:=median (B, N/ 2)    -   O (1)
By reducing the size of the arrays by half each iteration we logarithmically decreasing the running time and thus improving efficiency – O (logN)
And we repeat until size = 1 or 2


Efficiency :

T(N) = T(N/2) + 1
T(1) = 1
T(N) = T(N/2) + 1 = T(N/4) + 1 + 1 = .......
T(N/(2^k)+k)
N/(2^k) = 1
k = logN  so - > T(N) = logN + 1

We can sum the efficiency of the algorithm to O (logN)