Morphocompletion for #7855 ⟨a, b, c | ab=1, cbc=bcc⟩

Back:⟨a, b, c | ab=1, cbc=bcc⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ab ⇒ 1
2. bcc ⇒ cbc
3. acbc ⇒ cc
4. accbc ⇒ ccc
5. acccbc ⇒ cccc
6. accccbc ⇒ ccccc
7. acccccbc ⇒ cccccc
8. accccccbc ⇒ ccccccc
9. acccccccbc ⇒ cccccccc
10. accccccccbc ⇒ ccccccccc
11. acccccccccbc ⇒ cccccccccc
12. accccccccccbc ⇒ ccccccccccc
13. acccccccccccbc ⇒ cccccccccccc
14. accccccccccccbc ⇒ ccccccccccccc
15. acccccccccccccbc ⇒ cccccccccccccc
16. accccccccccccccbc ⇒ ccccccccccccccc
17. acccccccccccccccbc ⇒ cccccccccccccccc
18. accccccccccccccccbc ⇒ ccccccccccccccccc
19. acccccccccccccccccbc ⇒ cccccccccccccccccc
20. accccccccccccccccccbc ⇒ ccccccccccccccccccc
...

Collecting factors up to length 4 / frequency 7:

[2/0]cc155lf:154,ls:1,rf:171,rp:18,rs:18,re:1,a:1
[2/1]bc38lf:19,lp:1,ls:18,rf:1,rs:1,lprs:10,a:1
[2/2]ac36lf:18,lp:18,a:2
[2/3]cb18lf:18,rf:1,rp:1,a:1
[3/0]ccc136lf:136,rf:153,rp:17,rs:17,re:1,a:3
[3/1]cbc36lf:18,ls:18,rf:1,rp:1,rs:1,re:1,a:1
[3/2]acc34lf:17,lp:17,a:3
[3/3]ccb17lf:17,a:3
[3/4]acb2lf:1,lp:1,a:2

Considering [length 3 / frequency 1] cbc=d.

Step 2

Rewriting system is complete. See ⟨a, b, c | ab=1, cbc=bcc⟩.