Morphocompletion for #3536 ⟨a, b, c | bb=ac, caa=a⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ac ⇒ bb
2. cabb ⇒ bb
3. bbabb ⇒ abb
4. caa ⇒ a
5. bbaa ⇒ aa
6. cababb ⇒ babb
7. bbababb ⇒ ababb
8. cabaa ⇒ baa
9. bbabaa ⇒ abaa
10. cabababb ⇒ bababb
11. bbabababb ⇒ abababb
12. cababaa ⇒ babaa
13. bbababaa ⇒ ababaa
14. cababababb ⇒ babababb
15. bbababababb ⇒ ababababb
16. cabababaa ⇒ bababaa
17. bbabababaa ⇒ abababaa
18. cabababababb ⇒ bababababb
19. cababababaa ⇒ babababaa
20. bbababababaa ⇒ ababababaa
...

Collecting factors up to length 4 / frequency 7:

[2/0]ab45lf:45,rf:36,rp:8,a:3
[2/1]ba45lf:45,rf:36,rp:8,a:2
[2/2]bb36lf:18,lp:9,ls:9,rf:10,rp:2,rs:10,re:2,lsrp:1,lprs:4,lsrs:9
[2/3]aa20lf:10,ls:10,rf:9,rp:1,rs:9,re:1,lsrp:1,lsrs:9,a:1
[2/4]ca20lf:10,lp:10,a:1
[3/0]aba36lf:36,rf:28,rp:7,a:5
[3/1]bab36lf:36,rf:28,rp:7,a:4
[3/2]baa18lf:9,ls:9,rf:8,rp:1,rs:8,re:1,lsrp:1,lsrs:8,a:2
[3/3]abb18lf:9,ls:9,rf:8,rp:1,rs:8,re:1,lsrp:1,lsrs:8,a:3
[3/4]cab18lf:9,lp:9,a:3
[3/5]bba18lf:9,lp:9,a:2

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

Step 2

Rewriting system is complete. See ⟨a, b, c | bb=ac, caa=a⟩.