A tricky problem in Type Theory has a short satisfactory solution. To check if two potentially recursive expressions are equal, first we take the weak head normal form of both sides, without unrolling fixed points. Then, we do:
(F . G) == (G . F)
------------------- Fix=Fix case: apply T6's Theorem
μk(F k) == μk(G k)
(F Y) == Y
------------- Fix=Val case: apply T6's Theorem
μk(F k) == Y