First it’s a resistance loss, not a resistance loose. AFAIK.
second it’s literally -20% to the resists, so multiplication by 0.8
Then you need to understand the meaning of resist gain/loss. Each resist application, after stacking penalty, is literally a resonnance layer : it multiplies the damage incoming by (100-res)/100 .
If you have a module that provides a -10% resist, and then another one which does the same, it actually multiplies damage by 0.9 then 0.9 again, so total damages are multiplied by 0.81 : two -10% modules are the same as a -19% , or in other words a -19% resist module has the value of 2 resist 10 Equivalent (R10E)
Formally, the number of -10% module equivalent for a resist of -r (remember, again after stacking penalty) is R10E(r) = ln(1-r/100)/ln(1-10/100)
here is the graph, with x the resistance of a module after penalties, with y the number of effective -10% module that is the same value.
Now, the question is : what do we lose with that patch ? so the answer is, the difference : what we had before, minus what we will have after patch.
Formally, the loss 10% Equivalent (L10E) :
L10E(r) = ln(1-r/100)/ln(0.9) - ln(1-0.8r/100)/ln(0.9)
= ln((100-r)/(100-0.8r))/ln(0.9)
I add the value to the previous graph here and we can notice that the loss is of a -10% equivalent for approximately a -36% module (the loss of a -36% is approximately 1 R10E )
So yes, the loss of ONE module that was -36% can be compensated for by adding a -10% module somewhere else.
Now if you add two modules and one of them was stacking penalized, you need to add the loss of the first and the loss of the stacking penalized second.
Before the patch, two invul gave as much as :
( ln(1-0.3)+ln(1-0.3×0.869 ) )/ln(0.9) =6.2521 -10% resist equivalent.(R10E)
After the patch, two invuls will give as much as (assuming -26%) :
( ln(1-0.26)+ln(1-0.26×0.869 ) /ln(0.9) = 5.29 R10E
So yeah the addition of a -10% resist module does compensate for it, but that’s because the invuls are not subject to the -20%.
If the invuls were actually subject to the -20% they would become 4.83 R1OE so a loss of ± 1.43 R10E. so no, you could not compensate the loss over two -30% before penalty by a -10% module (you would need a -14% module for 0.9^1.43 = 0.86)
edit : forgot a parenthesis, everything was wrong, my b.