Morphocompletion for #1940 ⟨a, b | aabaaaba=ab⟩

Back:⟨a, b | aabaaaba=ab⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabaab ⇒ abaaba
2. aabaaaba ⇒ ab
3. ababaab ⇒ abbaaba
4. ababaaaba ⇒ abb
5. abaabaaaaba ⇒ aabab
6. aababab ⇒ abababa
7. aabaaabb ⇒ abbaaaba
8. abbabaab ⇒ abbbaaba
9. abbabaaaba ⇒ abbb
10. abababab ⇒ abbababa
11. aababbab ⇒ ababbaba
12. abbbabaaaba ⇒ abbbb
...

Collecting factors up to length 8 / frequency 7:

[2/0]ab47lf:34,lp:7,ls:6,rf:23,rp:11,rs:2,re:1,lprp:6,lsrp:6,lprs:1
[2/1]ba32lf:27,ls:5,rf:18,rs:7
[2/2]aa27lf:22,lp:5,rf:6,rp:1
[2/3]bb7lf:6,ls:1,rf:12,rs:3,a:2
[3/0]aba32lf:23,lp:4,ls:5,rf:13,rp:3,rs:7,lprs:2
[3/1]aab23lf:15,lp:5,ls:3,rf:5,rp:1
[3/2]bab15lf:12,ls:3,rf:7,rs:1,a:2
[3/3]baa10lf:10,rf:4
[3/4]abb9lf:5,lp:3,ls:1,rf:8,rp:7,rs:1,re:1,lprp:3,lsrp:1,a:2
[3/5]aaa7lf:7,rf:1
[3/6]bba4lf:4,rf:5,a:5
[4/0]aaba21lf:11,lp:5,ls:5,rf:5,rp:1,rs:4,lsrp:1,lprs:2
[4/1]abab13lf:8,lp:3,ls:2,rf:5,rp:2,rs:1,lsrp:1,a:2
[4/2]abaa11lf:10,lp:1,rf:1,rp:1
[4/3]baba8lf:8,rf:5,rs:3,a:2
[4/4]baab7lf:4,ls:3,rf:3,a:1
[4/5]aaab6lf:6,rf:1
[4/6]baaa6lf:6,rf:1
[5/0]aaaba10lf:5,ls:5,rf:1,rs:1
[5/1]abaab8lf:4,lp:1,ls:3,rf:1,rp:1,lsrp:10,a:1
[5/2]ababa8lf:5,lp:3,rf:3,rp:1,rs:2,lprs:1,a:2
[5/3]aabaa7lf:4,lp:3
[5/4]abbab6lf:3,lp:2,ls:1,rf:2,rp:1,a:6
[5/5]abaaa6lf:6
[5/6]babab5lf:3,ls:2,rf:2,a:20
[6/0]baaaba8lf:4,ls:4,rf:1,rs:1
[6/1]ababab6lf:3,lp:1,ls:2,rf:1,rp:1,lsrp:10,a:20
[6/2]abaaab5lf:5
[6/3]aabaaa5lf:3,lp:2
[6/4]abbaba4lf:2,lp:2,rf:2,rp:1,rs:1,a:6
[6/5]babaab4lf:2,ls:2,a:7
[6/6]ababaa4lf:2,lp:2,a:2
[7/0]abaaaba8lf:4,ls:4
[7/1]abbabaa4lf:2,lp:2,a:6
[7/2]aabaaab4lf:2,lp:2
[7/3]babaaab3lf:3,a:2
[7/4]ababbab2lf:1,ls:1,rf:1,rp:1,lsrp:10,a:24
[7/5]abbbaba2lf:1,lp:1,a:9
[7/6]bababab2lf:1,ls:1,a:27

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

Step 2

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