Morphocompletion for #954 ⟨a, b | aabbaba=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. aabbaba ⇒ ba
2. aabbabba ⇒ bba
3. aabbabbba ⇒ bbba
4. aabbabbbba ⇒ bbbba
5. aabbabbbbba ⇒ bbbbba
6. aabbabbbbbba ⇒ bbbbbba
7. aabbabbbbbbba ⇒ bbbbbbba
8. aabbabbbbbbbba ⇒ bbbbbbbba
9. aabbabbbbbbbbba ⇒ bbbbbbbbba
10. aabbabbbbbbbbbba ⇒ bbbbbbbbbba
11. aabbabbbbbbbbbbba ⇒ bbbbbbbbbbba
12. aabbabbbbbbbbbbbba ⇒ bbbbbbbbbbbba
13. aabbabbbbbbbbbbbbba ⇒ bbbbbbbbbbbbba
14. aabbabbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbba
15. aabbabbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbba
16. aabbabbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbba
17. aabbabbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbba
18. aabbabbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbba
19. aabbabbbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbbba
20. aabbabbbbbbbbbbbbbbbbbbbba ⇒ bbbbbbbbbbbbbbbbbbbba
...

Collecting factors up to length 8 / frequency 7:

[2/0]bb210lf:210,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]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: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]aabb40lf: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]babb19lf:19,a:1
[4/6]abbb18lf:18,a:2
[5/0]bbbbb136lf:136,rf:136,rp:16,a:4
[5/1]aabba40lf: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]bbabb19lf:19,a:1
[5/5]babbb18lf:18,a:2
[5/6]abbbb17lf:17,a:3
[6/0]bbbbbb120lf:120,rf:120,rp:15,a:5
[6/1]aabbab40lf: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]abbabb19lf:19,a:1
[6/4]bbabbb18lf:18,a:2
[6/5]babbbb17lf:17,a:3
[6/6]abbbbb16lf:16,a:4
[7/0]bbbbbbb105lf:105,rf:105,rp:14,a:6
[7/1]aabbabb38lf:19,lp:19,a:1
[7/2]bbbbbba30lf:15,ls:15,rf:15,rp:1,rs:15,re:1,lsrp:1,lsrs:15,a:5
[7/3]abbabbb18lf:18,a:2
[7/4]bbabbbb17lf:17,a:3
[7/5]babbbbb16lf:16,a:4
[7/6]abbbbbb15lf:15,a:5

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

Step 2

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