Inequality of Clausius I'm having some trouble understanding the inequality of Clausius. My treatment of the inequality applies to an engine operating in a cycle between two temperatures (i.e. like a Carnot engine). The inequality states that for an irreversible process:
$$ (1)\ \int _{closed\ path} \frac {dq_{irr}}{T}<0  $$
From what I've been reading, $ {dq_{irr}} $ is the heat added to the system (the engine) and $ T $ is the temperature of the reservoirs, $ T_{res} $. If this is the case, then $-{dq_{irr}}$ is the heat added to the reservoirs, $ dq_{irr,res} $, and
$$ (2)\  \int _{closed\ path} \frac {-dq_{irr}}{T_{res}}=\int _{closed\ path} \frac {dq_{irr,res}}{T_{res}}>0  $$
I read somewhere that if the reservoir is infinite then any heat added is actually occurring infinitely slowly. Is this true? If this is the case, then
$$ (3)\  dq_{irr,res} = dq_{rev, res} $$
$$ (4)\  \int _{closed\ path} \frac {dq_{rev, res}}{T_{res}}=∆S_{res}>0  $$
This seems reasonable. Since the engine itself is operating in a cycle, $ ∆S_{sys} = 0 $. This therefore satisfies the second law of thermodynamics:
$$ (5)\  ∆S_{univ}=∆S_{sys}+∆S_{res}=0+∆S_{rev}>0 $$
Does this all look right? $ (5) $ seems to say that the entropy change for an irreversible process is 0 for the case of the system itself, but is > 0 for the case of the surroundings. Is this only the case for the situation I have described?
I am also unsure about $ (3) $. If the heat is being added to the reservoirs infinitely slowly, then wouldn't the same be true of the system? This would make $ ∆S_{sys} = -∆S_{res} ≠ 0 $ and is obviously not correct.
 A: 
I read somewhere that if the reservoir is infinite then any heat added is actually occurring infinitely slowly. Is this true? 

No, infinite reservoir does not imply infinitely slow transfer of heat. There is no direct relation.
Equation (4) is incorrect; if the integral is over closed path in the state space of the reservoir, its value is 0. If the integral corresponds to cycle made on the system where something is irreversible (like heat transfer across finite temperature difference), in the end, when the system returns to its original equilibrium state and the reservoir to some equilibrium state, the entropy of the reservoir will be higher than before (in the case of the heat transfer across finite temperature difference entropy is generated at the interface of the two systems with gradient of temperature).
A: Exactly as you said: "(5) seems to say that the entropy change for an irreversible process is 0 for the case of the system itself, but is > 0 for the case of the surroundings." After the cycle is finished, by definition of the "cycle", the system returns to its original state irrespective of all the irreversibilities that may have taken place within it during some portions of the cycle. 
Entropy being a state variable it also assumes its initial value after the cycle. The only place therefore where the entropy will have increased after the cycle is in the reservoirs that do not perform any "cycle", they are sinks and sources of heat and their state can and will change per Clausius., i.e., $$\Delta S_{res} =\frac{q_{res}}{T_{res}}$$
As regards to equality vs. inequality I always learned and thought I understood that any time there is irreversible process it is inequality and never equality, and that goes for any part of the process. This can be understood intuitively by noting that irreversibility is caused by internal dissipation, i.e., heat generation, therefore during irreversible processes less heat is absorbed from the reservoir and more heat is rejected to the reservoir if compared to a reversible transition. 
Incidentally, the absorbed or rejected $q$ of a reservoir is always reversible, that is why its entropy change is just the ratio of the heat and its temperature.
