Morphocompletion for #3937 ⟨a, b | aaaabbaba=ba⟩

Solved by morph:3/2. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. aaaabbaba ⇒ ba
2. aaaabbabba ⇒ bba
3. aaaabbabbba ⇒ bbba
4. aaaabbabbbba ⇒ bbbba
5. aaaabbabbbbba ⇒ bbbbba
6. aaaabbabbbbbba ⇒ bbbbbba
7. aaaabbabbbbbbba ⇒ bbbbbbba
8. aaaabbabbbbbbbba ⇒ bbbbbbbba
9. aaaabbabbbbbbbbba ⇒ bbbbbbbbba
10. aaaabbabbbbbbbbbba ⇒ bbbbbbbbbba
11. aaaabbabbbbbbbbbbba ⇒ bbbbbbbbbbba
12. aaaabbabbbbbbbbbbbba ⇒ bbbbbbbbbbbba
13. aaaabbabbbbbbbbbbbbba ⇒ bbbbbbbbbbbbba
14. aaaabbabbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbba
15. aaaabbabbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbba
16. aaaabbabbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbba
17. aaaabbabbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbba
18. aaaabbabbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbba
19. aaaabbabbbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbbba
20. aaaabbabbbbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbbbba
...

Collecting factors up to length 8 / frequency 7:

[2/0]bb210lf:210,rf:190,rp:19
[2/1]aa80lf:60,lp:20
[2/2]ba60lf:40,ls:20,rf:20,rp:1,rs:20,re:1,lsrp:1,lsrs:20
[2/3]ab40lf:40
[3/0]bbb171lf:171,rf:171,rp:18,a:2
[3/1]aaa60lf:40,lp:20
[3/2]bba58lf:39,ls:19,rf:19,rp:1,rs:19,re:1,lsrp:1,lsrs:19
[3/3]abb39lf:39
[3/4]bab20lf:20
[3/5]aab20lf:20
[3/6]aba2lf:1,ls:1
[4/0]bbbb153lf:153,rf:153,rp:17,a:3
[4/1]aaaa40lf:20,lp:20
[4/2]bbba36lf:18,ls:18,rf:18,rp:1,rs:18,re:1,lsrp:1,lsrs:18,a:2
[4/3]abba22lf:21,ls:1
[4/4]bbab20lf:20
[4/5]aabb20lf:20
[4/6]aaab20lf:20
[5/0]bbbbb136lf:136,rf:136,rp:16,a:4
[5/1]aaaab40lf: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]abbab20lf:20
[5/4]aabba20lf:20
[5/5]aaabb20lf:20
[5/6]bbabb19lf:19,a:1
[6/0]bbbbbb120lf:120,rf:120,rp:15,a:5
[6/1]aaaabb40lf: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]aabbab20lf:20
[6/4]aaabba20lf:20
[6/5]abbabb19lf:19,a:1
[6/6]bbabbb18lf:18,a:2
[7/0]bbbbbbb105lf:105,rf:105,rp:14,a:6
[7/1]aaaabba40lf: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]aaabbab20lf:20
[7/4]aabbabb19lf:19,a:1
[7/5]abbabbb18lf:18,a:2
[7/6]bbabbbb17lf:17,a:3

Considering [length 3 / frequency 2] bba=c.

Step 2

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