Two permutations of the integers $1,2,3,...,n$ are given. The first is $p_1=(1,2,...,n)$ and the second is $p_2 = (i_1,i_2,...,i_n)$ .
(Operation) The following 'operation' can be applied on an arbitrary permutation $(j_1,j_2,...,j_n)$:
Take two adjacent numbers and swap them.
For example, if $n=5$ and one permutation of the numbers $1,2,...,5$ is $p_5=(4,1,5,3,2)$, then if we apply the 'operation' with this permutation, we can get: $(4,1,5,3,2) -> (4,1,3,5,2)$ - we swapped '$5$' and '$3$' and received $(4,1,3,5,2)$. Or, for example if we swap '$4$' and '$1$', we will receive $(1,4,5,3,2)$. (End of Operation)
(Transformation Method) If we try to transform the given permutation $(i_1,i_2,...,i_n)$ in $(1,2,...,n)$ using the given 'operation', we can do the following: find the number '$1$' in $(i_1,i_2,...,i_n)$ and apply the 'operation' on the number '$1$' until we reach $pp = (1,-,-,...,-) $, after this find the number '$2$' in $pp$ and apply the operation on '$2$' until we reach $ppp= (1,2,-,...,-)$ and so on for every number from $1$ to $n$. (End of Transformation Method)
Of course , other ways for transforming $(i_1,i_2,...,i_n)$ into $(1,2,...,n)$ exist , which are different from the suggested method (Operation above) and these other methods can use a bigger number of 'operation' (a bigger number swaps) in order to achieve $(1,2,...,n)$.
Question: How can I prove, that the suggested method above for transforming $(i_1,i_2,...,i_n)$ into $(1,2,...,n)$ is doing the minimum number of 'operation'?
That is - if we transform $(i_1,i_2,...,i_n)$ into $(1,2,...,n)$ using the 'operation' in an arbitrary way then the number of used 'operation' is $\ge$ than the number of operations used by the described method (Transformation Method above).
Thanks