Morphocompletion for #6101 ⟨a, b, c | ab=c, cac=ba⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ab ⇒ c
2. cac ⇒ ba
3. baac ⇒ cca
4. acca ⇒ caac
5. caacb ⇒ accc
6. caacc ⇒ acba
7. ccaac ⇒ baca
8. acbaa ⇒ baaac
9. baaacb ⇒ acbac
10. baaacc ⇒ accca
11. aacbac ⇒ caaacb
12. bacaac ⇒ cbaa
13. cbaaac ⇒ bcaa
14. bcaaac ⇒ cccaa
15. aaccca ⇒ caaacc
16. acccaa ⇒ ccaaac
17. caaaccb ⇒ aacccc
18. ccaaacb ⇒ accba
19. baaaacb ⇒ aacbca
20. caaaccc ⇒ aaccba
...

Collecting factors up to length 4 / frequency 7:

[2/0]ac31lf:21,lp:3,ls:7,rf:15,rp:5,rs:4,lprs:3,a:1
[2/1]aa30lf:26,lp:2,ls:2,rf:15,rp:3,rs:3,a:2
[2/2]ca19lf:12,lp:5,ls:2,rf:10,rp:3,rs:4,lsrp:2,a:1
[2/3]cc17lf:12,lp:2,ls:3,rf:14,rp:3,rs:3,a:2
[2/4]cb14lf:8,lp:1,ls:5,rf:7,rp:1,rs:1,a:5
[2/5]ba13lf:8,lp:5,rf:8,rp:3,rs:4,re:1,a:1
[2/6]bc2lf:1,lp:1,rf:2,rp:1,a:15
[3/0]aac22lf:15,lp:2,ls:5,rf:8,rp:3,rs:3,a:2
[3/1]caa14lf:9,lp:4,ls:1,rf:6,rp:3,rs:2,a:3
[3/2]baa11lf:6,lp:4,ls:1,rf:2,rp:1,rs:1,lsrp:1,a:2
[3/3]acc11lf:7,lp:2,ls:2,rf:6,rp:3,rs:1,lsrp:1,a:3
[3/4]acb11lf:6,lp:1,ls:4,rf:4,rp:2,rs:1,lsrp:1,a:6
[3/5]aaa9lf:9,rf:4,a:8
[3/6]cca9lf:5,lp:2,ls:2,rf:4,rp:2,rs:2,re:1,a:2

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

Step 2

Rewriting system is complete. See ⟨a, b, c | ab=c, cac=ba⟩.