One of the finer points of amortized analysis about which I have been able to find relatively little information is the broad question of what happens to the amortized cost of a structure's existing operations if we add a new one. I'll use two popular examples to illustrate my confusion:
In a binary counter which only supports the increment operation, choosing the potential function $\Phi=\#$ of $1$'s in the present configuration gives us constant amortized cost for the operation. In every example I can see in the literature, adding the decrement operation to the counter creates a situation where both it and the increment operation have linear amortized costs. The argument given is that we can toggle $m$ times between values of $2^n$ and $2^n-1$ with a series of alternating decrement and increment operations, giving us a time of $\Theta(mn)$.
In a binomial heap, using the number of binomial trees in the structure as a potential function gives us constant amortized time for insertion. Popping the minimum takes logarithmic time both amortized and in actuality.
Perhaps I'm mistaken, but I see a certain contradiction here. In a binomial heap we could reach a point where we have one large binomial tree with $2^n$ elements and repeatedly perform pop and push operations, which will cost $\Theta(m\log n)$ time. Even if each vertex keeps its children in a linked list which would give the deletions constant time, the insertions will all take logarithmic time.
Having said that, the insertion operations for both structures still function the same way: for any sufficiently large $m$, any series of $m$ insertions to either structure still takes $O(m)$ time - the inclusion of the deletion (decrement in the counter) operation doesn't affect the way the insertion operation works.
So why do we say that adding decrementation to a binary counter increases the amortized cost for incrementation and then ignore the same argument when analyzing binomial heaps?