Morphocompletion for #4016 ⟨a, b | aaabbaaba=ba⟩

Back:⟨a, b | aaabbaaba=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. aaabbaaba ⇒ ba
2. aaabbaabba ⇒ bba
3. aaabbaabbba ⇒ bbba
4. aaabbaabbbba ⇒ bbbba
5. aaabbaabbbbba ⇒ bbbbba
6. aaabbaabbbbbba ⇒ bbbbbba
7. aaabbaabbbbbbba ⇒ bbbbbbba
8. aaabbaabbbbbbbba ⇒ bbbbbbbba
9. aaabbaabbbbbbbbba ⇒ bbbbbbbbba
10. aaabbaabbbbbbbbbba ⇒ bbbbbbbbbba
11. aaabbaabbbbbbbbbbba ⇒ bbbbbbbbbbba
12. aaabbaabbbbbbbbbbbba ⇒ bbbbbbbbbbbba
13. aaabbaabbbbbbbbbbbbba ⇒ bbbbbbbbbbbbba
14. aaabbaabbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbba
15. aaabbaabbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbba
16. aaabbaabbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbba
17. aaabbaabbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbba
18. aaabbaabbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbba
19. aaabbaabbbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbbba
20. aaabbaabbbbbbbbbbbbbbbbbbbba ⇒ 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]bba58lf:39,ls:19,rf:19,rp:1,rs:19,re:1,lsrp:1,lsrs:19
[3/2]aab40lf:40
[3/3]aaa40lf:20,lp:20
[3/4]abb39lf:39
[3/5]baa20lf:20
[3/6]aba2lf:1,ls:1
[4/0]bbbb153lf:153,rf:153,rp:17,a:3
[4/1]aaab40lf:20,lp:20
[4/2]aabb39lf:39
[4/3]bbba36lf:18,ls:18,rf:18,rp:1,rs:18,re:1,lsrp:1,lsrs:18,a:2
[4/4]abba22lf:21,ls:1
[4/5]bbaa20lf:20
[4/6]baab20lf:20
[5/0]bbbbb136lf:136,rf:136,rp:16,a:4
[5/1]aaabb40lf: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]aabba22lf:21,ls:1
[5/4]bbaab20lf:20
[5/5]abbaa20lf:20
[5/6]baabb19lf:19,a:1
[6/0]bbbbbb120lf:120,rf:120,rp:15,a:5
[6/1]aaabba40lf: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]abbaab20lf:20
[6/4]aabbaa20lf:20
[6/5]bbaabb19lf:19,a:1
[6/6]baabbb18lf:18,a:2
[7/0]bbbbbbb105lf:105,rf:105,rp:14,a:6
[7/1]aaabbaa40lf: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]aabbaab20lf:20
[7/4]abbaabb19lf:19,a:1
[7/5]bbaabbb18lf:18,a:2
[7/6]baabbbb17lf:17,a:3

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

Step 2

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