Morphocompletion for #6229 ⟨a, b, c | aa=1, baccab=1⟩

Back:⟨a, b, c | aa=1, baccab=1⟩

Solved by morph:3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aa ⇒ 1
2. bbac ⇒ acabba
3. ccabb ⇒ a
4. bacc ⇒ accaba
5. ccabacabb ⇒ abaca
6. ccabacabacabb ⇒ abacabaca
7. ccabacabacabba ⇒ abacabac
8. ccabacabacabacabb ⇒ abacabacabaca
9. ccabacabacabacabba ⇒ abacabacabac
10. ccabacabacabacabacabb ⇒ abacabacabacabaca
11. ccabacabacabacabacabba ⇒ abacabacabacabac
12. ccabacabacabacabacabacabb ⇒ abacabacabacabacabaca
13. ccabacabacabacabacabacabba ⇒ abacabacabacabacabac
14. ccabacabacabacabacabacabacabb ⇒ abacabacabacabacabacabaca
15. ccabacabacabacabacabacabacabba ⇒ abacabacabacabacabacabac
...

Collecting factors up to length 4 / frequency 7:

[2/0]ba54lf:48,lp:1,ls:5,rf:43,rs:2,lprs:1,a:5
[2/1]ca53lf:53,rf:38,rs:6,a:5
[2/2]ab53lf:53,rf:43,rp:11,a:5
[2/3]ac44lf:43,ls:1,rf:43,rp:2,rs:5,lsrp:1,a:5
[2/4]cc26lf:13,lp:12,ls:1,rf:1,a:5
[2/5]bb21lf:13,lp:1,ls:7,rf:1,a:6
[3/0]cab53lf:53,rf:32,a:5
[3/1]bac45lf:43,lp:1,ls:1,rf:41,rs:5,a:5
[3/2]aca41lf:41,rf:37,rp:1,rs:6,a:8
[3/3]aba41lf:41,rf:42,rp:11,rs:1,a:5
[3/4]cca24lf:12,lp:12,rf:1,a:5
[3/5]abb19lf:12,ls:7,rf:1,a:6
[3/6]bba12lf:6,lp:1,ls:5,rf:1,rs:1,lprs:1,a:11

Considering [length 3 / frequency 0] cab=d.

Step 2

Rewriting system is complete. See ⟨a, b, c | aa=1, baccab=1⟩.