Morphocompletion for #2006 ⟨a, b | aabbbaba=ba⟩

Solved by morph:3/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aabbbaba ⇒ ba
2. aabbbabba ⇒ bba
3. aabbbabbba ⇒ bbba
4. aabbbabbbba ⇒ bbbba
5. aabbbabbbbba ⇒ bbbbba
6. aabbbabbbbbba ⇒ bbbbbba
7. aabbbabbbbbbba ⇒ bbbbbbba
8. aabbbabbbbbbbba ⇒ bbbbbbbba
9. aabbbabbbbbbbbba ⇒ bbbbbbbbba
10. aabbbabbbbbbbbbba ⇒ bbbbbbbbbba
11. aabbbabbbbbbbbbbba ⇒ bbbbbbbbbbba
12. aabbbabbbbbbbbbbbba ⇒ bbbbbbbbbbbba
13. aabbbabbbbbbbbbbbbba ⇒ bbbbbbbbbbbbba
14. aabbbabbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbba
15. aabbbabbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbba
16. aabbbabbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbba
17. aabbbabbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbba
18. aabbbabbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbba
19. aabbbabbbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbbba
20. aabbbabbbbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbbbba
...

Collecting factors up to length 8 / frequency 7:

[2/0]bb230lf:230,rf:190,rp:19
[2/1]ba60lf:40,ls:20,rf:20,rp:1,rs:20,re:1,lsrp:1,lsrs:20
[2/2]ab40lf:40
[2/3]aa40lf:20,lp:20
[3/0]bbb191lf:191,rf:171,rp:18
[3/1]bba58lf:39,ls:19,rf:19,rp:1,rs:19,re:1,lsrp:1,lsrs:19
[3/2]aab40lf:20,lp:20
[3/3]abb39lf:39
[3/4]bab20lf:20
[3/5]aba2lf:1,ls:1
[4/0]bbbb153lf:153,rf:153,rp:17,a:3
[4/1]bbba56lf:38,ls:18,rf:18,rp:1,rs:18,re:1,lsrp:1,lsrs:18
[4/2]aabb40lf:20,lp:20
[4/3]abbb38lf:38
[4/4]bbab20lf:20
[4/5]babb19lf:19,a:1
[4/6]abba2lf:1,ls:1,a:1
[5/0]bbbbb136lf:136,rf:136,rp:16,a:4
[5/1]aabbb40lf:20,lp:20
[5/2]bbbba34lf:17,ls:17,rf:17,rp:1,rs:17,re:1,lsrp:1,lsrs:17,a:3
[5/3]abbba22lf:21,ls:1
[5/4]bbbab20lf:20
[5/5]bbabb19lf:19,a:1
[5/6]babbb18lf:18,a:2
[6/0]bbbbbb120lf:120,rf:120,rp:15,a:5
[6/1]aabbba40lf:20,lp:20
[6/2]bbbbba32lf:16,ls:16,rf:16,rp:1,rs:16,re:1,lsrp:1,lsrs:16,a:4
[6/3]abbbab20lf:20
[6/4]bbbabb19lf:19,a:1
[6/5]bbabbb18lf:18,a:2
[6/6]babbbb17lf:17,a:3
[7/0]bbbbbbb105lf:105,rf:105,rp:14,a:6
[7/1]aabbbab40lf:20,lp:20
[7/2]bbbbbba30lf:15,ls:15,rf:15,rp:1,rs:15,re:1,lsrp:1,lsrs:15,a:5
[7/3]abbbabb19lf:19,a:1
[7/4]bbbabbb18lf:18,a:2
[7/5]bbabbbb17lf:17,a:3
[7/6]babbbbb16lf:16,a:4

Considering [length 3 / frequency 1] bba=c.

Step 2

Rewriting system is complete. See ⟨a, b | aabbbaba=ba⟩.