Whoever did this has never drawn a swastika before. The one on the right is backwards, and the left one is ????https://twitter.com/NYDailyNews/status/799766392383930368 …
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A false flag false flag is a false flag with a false flag, but a false false flag is a false false flag, t
Sorry I meant to say (Fa Fl)^2 ≡ Fa (Fa Fl) => (Fa Fl)^n ≡ Fa^{n-1} (Fa Fl) ≡ Fa^n Fl
But (false flag = flag false) => flag is idempotent => fl=A⊕B, where A,B invariant on False and not False
That's only true if you have commutativity and inverses
IIRC, You only need [fa, fl] = 0 (proved) and False nonsingular (which I assumed was obvious)
As a word problem [fa, fl] = 0 is not proven, we don't even know inverses exist.
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