Morphocompletion for #2005 ⟨a, b | aabbbaba=ab⟩

Solved by morph:4/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabbbaba ⇒ ab
2. ababbbaba ⇒ abb
3. aabbbabb ⇒ abbbbaba
4. abbabbbaba ⇒ abbb
5. ababbbabb ⇒ abbbbbaba
6. abbbbabababa ⇒ aabbbb
7. abbbabbbaba ⇒ abbbb
8. abbabbbabb ⇒ abbbbbbaba
9. abbbbbabababa ⇒ ababbbb
10. abbbbabbbaba ⇒ abbbbb
11. abbbabbbabb ⇒ abbbbbbbaba
12. aabbbbbbbaba ⇒ abbbbabababb
13. abbbbbbabababa ⇒ abbabbbb
14. abbbbbabbbaba ⇒ abbbbbb
15. abbbbabbbabb ⇒ abbbbbbbbaba
16. abbbbbbabbbaba ⇒ abbbbbbb
...

Collecting factors up to length 8 / frequency 7:

[2/0]bb73lf:68,ls:5,rf:60,rs:10
[2/1]ab61lf:48,lp:13,rf:26,rp:15,rs:1,re:1,lprp:12
[2/2]ba54lf:43,ls:11,rf:15,rs:5
[2/3]aa6lf:3,lp:3,rf:1,rp:1
[3/0]abb45lf:29,lp:11,ls:5,rf:17,rp:13,rs:2,re:1,lprp:9,lsrp:5
[3/1]bbb39lf:39,rf:43,rs:8
[3/2]bab32lf:32,rf:10
[3/3]aba32lf:19,lp:2,ls:11,rf:8,rp:1,rs:5,lsrp:1,lprs:1
[3/4]bba24lf:24,rf:7
[3/5]aab6lf:3,lp:3,rf:1,rp:1
[4/0]abbb31lf:22,lp:9,rf:14,rp:11,rs:1,re:1,lprp:6
[4/1]baba28lf:17,ls:11,rf:7,rs:5
[4/2]bbab24lf:24,rf:7
[4/3]bbba22lf:22,rf:6
[4/4]babb20lf:15,ls:5,rf:3,rs:1,a:1
[4/5]bbbb17lf:17,rf:29,rs:7,a:2
[4/6]abab10lf:8,lp:2,rf:3,rp:1,a:1
[5/0]bbbab22lf:22,rf:6
[5/1]bbaba19lf:11,ls:8,rf:6,rs:5
[5/2]bbabb18lf:13,ls:5,rf:1,a:2
[5/3]abbba16lf:14,lp:2
[5/4]abbbb15lf:8,lp:7,rf:13,rp:10,rs:4,re:1,lprp:4,lprs:3,a:2
[5/5]babbb10lf:10,rf:2,a:1
[5/6]bbbbb9lf:9,rf:16,rs:3,a:4
[6/0]bbbaba19lf:11,ls:8,rf:6,rs:5
[6/1]bbbabb16lf:11,ls:5,a:2
[6/2]abbbab16lf:14,lp:2
[6/3]babbba10lf:10,a:1
[6/4]abbbbb9lf:5,lp:4,rf:7,rp:7,rs:1,re:1,lprp:2,a:4
[6/5]bababa9lf:6,ls:3,rf:1,a:10
[6/6]bbabbb8lf:8,rf:1,a:3
[7/0]abbbabb14lf:7,lp:2,ls:5,a:2
[7/1]abbbaba14lf:7,ls:7
[7/2]babbbab10lf:10,a:1
[7/3]bbabbba8lf:8,a:3
[7/4]bbbabbb6lf:6,a:5
[7/5]abbbbab6lf:3,lp:3,rf:2,rp:2,a:2
[7/6]abababa6lf:3,ls:3,a:10

Considering [length 4 / frequency 0] abbb=c.

Step 2

Rewriting system is complete. See ⟨a, b | aabbbaba=ab⟩.