Pick a Carnot Cycle (being $T_1<T_2$), it is reversible, therefore $\Delta S_{univ, cycle}=0$.

The same result is obtained via the sum of all entropies associated with its transformation, which means:
$\Delta S_{univ, cycle} = \Delta S_{gas+ambient,AB} + \Delta S_{gas+ambient,BC}+\Delta S_{gas+ambient,CD} +\Delta S_{gas+ambient,DA}$

Make the first adiabatic expansion irreversible, like in the picture:
[![enter image description here][1]][1]

The same equation applies, however, since entropy is a state function, $S_{gas, cycle}=0$, 
$\Delta S_{univ, cycle} = \Delta S_{ambient,AB} + \Delta S_{ambient,BC}+\Delta S_{ambient,CD} +\Delta S_{ambient,DA}$

The adiabatic processes, BC and DA don't account for a change in entropy, as $Q_{exchanged} =0$, so $\Delta S_{amb,BC} = \int\limits_{B}^{C} \frac {dQ} T$ goes to $0$, same for DA

Therefore the total entropy gets to $\Delta S_{univ, cycle} = \Delta S_{ambient,AB} + \Delta S_{ambient,CD} = \Delta S_{univ, irreversible processes} = S_{gas+ambient,BC} = S_{gas,BC} $

How can the entropy **also** become not dependent on the irreversible process? *Since it only depends on the isothermal transformations, how can different "degrees of irreversibility in BC" not affect the net entropy change in the universe?*

PS: does this relate to the fact that entropy is defined as function of state of any reversible process? It always comes up like magic in my calculations and I can't explain why.

EDIT: Possibly a duplicate of [this][2], but I'd actually prefer to know how this relates to cycle-wide calculations


  [1]: https://i.sstatic.net/T97Sk.png
  [2]: https://physics.stackexchange.com/questions/149729/entropy-change-in-an-irreversible-process?rq=1