Description: Lemma for kur14 : main proof. The set T here contains all the distinct combinations of k and c that can arise, and we prove here that applying k or c to any element of T yields another elemnt of T . In operator shorthand, we have T = { A , c A , k A , c k A , k c A , c k c A , k c k A , c k c k A , k c k c A , c k c k c A , k c k c k A , c k c k c k A , k c k c k c A , c k c k c k c A } . From the identities c c A = A and k k A = k A , we can reduce any operator combination containing two adjacent identical operators, which is why the list only contains alternating sequences. The reason the sequences don't keep going after a certain point is due to the identity k c k A = k c k c k c k A , proved in kur14lem6 . (Contributed by Mario Carneiro, 11-Feb-2015)
Ref | Expression | ||
---|---|---|---|
Hypotheses | kur14lem.j | |
|
kur14lem.x | |
||
kur14lem.k | |
||
kur14lem.i | |
||
kur14lem.a | |
||
kur14lem.b | |
||
kur14lem.c | |
||
kur14lem.d | |
||
kur14lem.t | |
||
Assertion | kur14lem7 | |