Morphocompletion for #3424 ⟨a, b, c | ba=ab, aca=c⟩

Solved by morph:2/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ba ⇒ ab
2. aca ⇒ c
3. abca ⇒ bc
4. abbca ⇒ bbc
5. abbbca ⇒ bbbc
6. abbbbca ⇒ bbbbc
7. abbbbbca ⇒ bbbbbc
8. acc ⇒ cca
9. acbc ⇒ cbca
10. acbbc ⇒ cbbca
11. acbbbc ⇒ cbbbca
12. acbbbbc ⇒ cbbbbca
13. abcc ⇒ bcca
14. abcbc ⇒ bcbca
15. abcbbc ⇒ bcbbca
16. abcbbbc ⇒ bcbbbca
17. abbcc ⇒ bbcca
18. abbcbc ⇒ bbcbca
19. abbcbbc ⇒ bbcbbca
20. abbbcc ⇒ bbbcca
...

Collecting factors up to length 4 / frequency 7:

[2/0]bc31lf:22,ls:9,rf:22,rp:5,rs:5,re:1,lsrp:3,a:2
[2/1]ab26lf:13,lp:13,rf:1,rp:1,rs:1,re:1
[2/2]bb25lf:25,rf:25,rp:8,a:4
[2/3]ac12lf:6,lp:6,a:1
[2/4]ca12lf:6,ls:6,rf:13,rs:13,a:1
[2/5]cb9lf:9,rf:9,rp:4,a:5
[2/6]cc8lf:4,ls:4,rf:4,rp:1,lsrp:10,a:3
[3/0]bbc20lf:14,ls:6,rf:14,rp:4,rs:4,re:1,lsrp:1,a:4
[3/1]abb16lf:8,lp:8,a:4
[3/2]bbb11lf:11,rf:11,rp:4,a:8
[3/3]abc10lf:5,lp:5,a:2
[3/4]bca10lf:5,ls:5,rf:9,rs:9,a:2
[3/5]acb8lf:4,lp:4,a:5
[3/6]bcc6lf:3,ls:3,rf:3,rp:1,lsrp:10,a:6

Considering [length 2 / frequency 1] ab=d.

Step 2

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