- #36
selfAdjoint
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AH! O-O -> "PHI" with one circle with the bar going vertically, which is of course the original graph. Neat! Now the three-specs. On the left two lenses reduce them to a "PHI" using your method. Now we have a PHI linked to a circle by a horizontal bar, The two nodes of the PHI plus the node at its end of the bar make a triangle so do a compression, which reduces the left PHI to just a circle joined to the bar. Now you have two-specs as before. So you have an induction for n-specs; given n=2 the problem is solved, and giving that it is solved for n=k-1, this shows it is also solved for n=k. So it is solved for all n [tex]\ge[/tex] 2.