User blog comment:Scorcher007/Analysis DAN up to Z2/@comment-11227630-20180919020025/@comment-31580368-20180919090559

Yes, this table is incorrect, since I did it, I learned a little more about stable ordinals.

according to Taranovsky and Rathjen

П12-CA0, which has the same strength as KP + {there is a Σ1 elementary chain of length a: a<ω}. eqiv. ω-ly (+1)-stable П12-CA, which has the same strength as KP + {there is a Σ1 elementary chain of length a: a<ω+1}. eqiv. ω+1-ly (+1)-stable Δ13-CA, which has the same strength as KP + {there is a Σ1 elementary chain of length a: a<ε0}. eqiv. ε0-ly (+1)-stable

In addition, this table also contains inaccuracies, I'm now rewriting it. I suppose that

S[a+2] ~ {n,n(1(1',,1,2,,)2)2} S[a×2] ~ {n,n(1(1',,1',,2,,)2)2} S[a2] ~ {n,n(1(1',,1',,1',,2,,)2)2} S[aa] ~ {n,n(1(1(2',,)2,,)2)2} this is still additionally S[aω+1] ~ {n,n(1,,,1,2)2} ~ П12-CA0 S[aΩ Ω Ω ... +1] ~ {n,n(1,,,1,,2)2 S[aα+1] ~ {n,n(1,,,1,,,2)2} ~ П12-TR0 S2[aω+1] ~ {n,n(1,,,,1,2)2} ~ П13-CA0 S2[aΩ Ω Ω ... +1] ~ {n,n(1,,,,1,,2)2} S2[aα+1] ~ {n,n(1,,,,1,,,2)2} S2[S[aα+1]-aα+1] ~ {n,n(1,,,,1,,,,2)2} ~ П13-TR0 S3[aω+1] ~ {n,n(1,,,,,1,2)2} ~ П14-CA0 S3[aΩ Ω Ω ... +1] ~ {n,n(1,,,,,1,,2)2} S3[aα+1] ~ {n,n(1,,,,,1,,,2)2} S3[S[aα+1]-aα+1] ~ {n,n(1,,,,,1,,,,2)2} S3[S2[S[aα+1]-aα+1]-aα+1] ~ {n,n(1,,,,,1,,,,,2)2} ~ П14-TR0 e.t.c.