No - there is a misunderstanding on your side.Using the schematic shown, when I apply a 1 amp test current to the non-inverting input of OP1 and solve for the voltage at that terminal, I don't get a conductance of (Y2 Y4 Y6)/Y3 Y5; my result involves Y0 and Y1, which surely must be true. Is the expression you give for Yin1 for a simpler circuit than shown in this post? I also get a different result for Yin2 for the circuit shown here.
In post#16 I wrote: "From known (basic) GIC chatacteristics, the input conductances at both non-inv. input nodes for OP1 and OP2 are....".
The well-known GIC-topology (introduced by Antoniou) consists of two opamps and five external impedances (resp. conductances), one of them grounded.
That means: The passive part(s) which are connected to the inv. input opamp node under consideration are, of course, NOT part of the corresponding input impedance/conductance.
This should be clear also because these external elements are part of the described voltage divider circuits - resulting in the voltages at the opamp input nodes Vp(1) and Vp(2), respectively (see my post #16).
(I hope my correction - change of "transconductance" into "connductance" - now was succesfull)