Morphocompletion for #4414 ⟨a, b, c | aba=1, acbc=c⟩

Back:⟨a, b, c | aba=1, acbc=c⟩

Solved by morph:2/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ba ⇒ ab
2. aab ⇒ 1
3. abc ⇒ cbc
4. acbc ⇒ c
5. abbc ⇒ cbbc
6. bcbc ⇒ cbbc
7. ccbbc ⇒ bc
8. acbbc ⇒ bc
9. abbbc ⇒ cbbbc
10. bcbbc ⇒ cbbbc
11. ccbbbc ⇒ bbc
12. acbbbc ⇒ bbc
13. abbbbc ⇒ cbbbbc
14. bcbbbc ⇒ cbbbbc
15. ccbbbbc ⇒ bbbc
16. acbbbbc ⇒ bbbc
17. ccbbbbbc ⇒ bbbbc
18. acbbbbbc ⇒ bbbbc
...

Collecting factors up to length 4 / frequency 7:

[2/0]bc38lf:19,lp:3,ls:16,rf:15,rp:2,rs:15,re:2,lsrp:2,lprs:3,lsrs:15,a:1
[2/1]bb29lf:29,rf:24,rp:6,a:5
[2/2]cb12lf:12,rf:7,rp:7,a:1
[2/3]ab10lf:5,lp:4,ls:1,rf:1,rp:1,rs:1,re:1,a:2
[2/4]ac10lf:5,lp:5,a:1
[2/5]cc8lf:4,lp:4,a:14
[2/6]aa2lf:1,lp:1,a:22
[3/0]bbc26lf:13,ls:13,rf:12,rp:2,rs:12,re:2,lsrp:2,lsrs:11,a:5
[3/1]bbb16lf:16,rf:12,rp:4,a:10
[3/2]acb10lf:5,lp:5,a:1
[3/3]cbb10lf:10,rf:6,rp:6,a:5
[3/4]ccb8lf:4,lp:4,a:14
[3/5]bcb6lf:3,lp:3,a:5
[3/6]abb6lf:3,lp:3,a:7

Considering [length 2 / frequency 0] bc=d.

Step 2

Rewriting system is complete. See ⟨a, b, c | aba=1, acbc=c⟩.