Morphocompletion for #4032 ⟨a, b | aaabbabba=ba⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. aaabbabba ⇒ ba
2. aaabbabbba ⇒ bba
3. aaabbabbbba ⇒ bbba
4. aaabbabbbbba ⇒ bbbba
5. aaabbabbbbbba ⇒ bbbbba
6. aaabbabbbbbbba ⇒ bbbbbba
7. aaabbabbbbbbbba ⇒ bbbbbbba
8. aaabbabbbbbbbbba ⇒ bbbbbbbba
9. aaabbabbbbbbbbbba ⇒ bbbbbbbbba
10. aaabbabbbbbbbbbbba ⇒ bbbbbbbbbba
11. aaabbabbbbbbbbbbbba ⇒ bbbbbbbbbbba
12. aaabbabbbbbbbbbbbbba ⇒ bbbbbbbbbbbba
13. aaabbabbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbba
14. aaabbabbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbba
15. aaabbabbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbba
16. aaabbabbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbba
17. aaabbabbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbba
18. aaabbabbbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbba
19. aaabbabbbbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbbba
20. aaabbabbbbbbbbbbbbbbbbbbbbba ⇒ 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]aa60lf:40,lp:20
[2/3]ab40lf:40
[3/0]bbb190lf:190,rf:171,rp:18,a:1
[3/1]bba60lf:40,ls:20,rf:19,rp:1,rs:19,re:1,lsrp:1,lsrs:19
[3/2]abb40lf:40
[3/3]aaa40lf:20,lp:20
[3/4]bab20lf:20
[3/5]aab20lf:20
[4/0]bbbb171lf:171,rf:153,rp:17,a:2
[4/1]aaab40lf:20,lp:20
[4/2]bbba38lf:19,ls:19,rf:18,rp:1,rs:18,re:1,lsrp:1,lsrs:18,a:1
[4/3]abba22lf:21,ls:1
[4/4]babb20lf:20
[4/5]bbab20lf:20
[4/6]aabb20lf:20
[5/0]bbbbb153lf:153,rf:136,rp:16,a:3
[5/1]aaabb40lf:20,lp:20
[5/2]bbbba36lf:18,ls:18,rf:17,rp:1,rs:17,re:1,lsrp:1,lsrs:17,a:2
[5/3]bbabb20lf:20
[5/4]abbab20lf:20
[5/5]aabba20lf:20
[5/6]babbb19lf:19,a:1
[6/0]bbbbbb136lf:136,rf:120,rp:15,a:4
[6/1]aaabba40lf:20,lp:20
[6/2]bbbbba34lf:17,ls:17,rf:16,rp:1,rs:16,re:1,lsrp:1,lsrs:16,a:3
[6/3]abbabb20lf:20
[6/4]aabbab20lf:20
[6/5]bbabbb19lf:19,a:1
[6/6]babbbb18lf:18,a:2
[7/0]bbbbbbb120lf:120,rf:105,rp:14,a:5
[7/1]aaabbab40lf:20,lp:20
[7/2]bbbbbba32lf:16,ls:16,rf:15,rp:1,rs:15,re:1,lsrp:1,lsrs:15,a:4
[7/3]aabbabb20lf:20
[7/4]abbabbb19lf:19,a:1
[7/5]bbabbbb18lf:18,a:2
[7/6]babbbbb17lf:17,a:3

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

Step 2

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