Morphocompletion for #2187 ⟨a, b | aaabbaa=baa⟩

Back:⟨a, b | aaabbaa=baa⟩

Solved by morph:5/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aaabbaa ⇒ baa
2. aaabbbaa ⇒ bbaa
3. aaabbabaa ⇒ babaa
4. aaabbbbaa ⇒ bbbaa
5. aaabbabbaa ⇒ babbaa
6. aaabbbabaa ⇒ bbabaa
7. aaabbbbbaa ⇒ bbbbaa
...

Collecting factors up to length 8 / frequency 7:

[2/0]aa35lf:21,lp:7,ls:7,rf:7,rs:7,lprs:7,lsrs:7
[2/1]bb15lf:15,rf:8,rp:4
[2/2]ba10lf:10,rf:10,rp:3
[2/3]ab10lf:10,rf:3
[3/0]baa14lf:7,ls:7,rf:7,rp:1,rs:7,re:1,lsrp:1,lsrs:7
[3/1]aaa14lf:7,lp:7
[3/2]bba8lf:8,rf:5,rp:2
[3/3]abb8lf:8,rf:1
[3/4]bbb7lf:7,rf:3,rp:2,a:1
[3/5]aab7lf:7
[3/6]bab3lf:3,rf:3,rp:2,a:2
[4/0]aaab14lf:7,lp:7
[4/1]bbaa10lf:5,ls:5,rf:4,rp:1,rs:4,re:1,lsrp:1,lsrs:4
[4/2]aabb7lf:7
[4/3]bbba4lf:4,rf:2,rp:1,a:1
[4/4]abbb4lf:4,a:1
[4/5]abba4lf:4,rf:1
[4/6]abaa4lf:2,ls:2,rf:2,rs:2,lsrs:2,a:2
[5/0]aaabb14lf:7,lp:7
[5/1]bbbaa6lf:3,ls:3,rf:2,rp:1,rs:2,re:1,lsrp:1,lsrs:2,a:1
[5/2]babaa4lf:2,ls:2,rf:2,rp:1,rs:2,re:1,lsrp:1,lsrs:2,a:2
[5/3]abbaa4lf:2,ls:2,rf:1,rs:1,lsrs:1
[5/4]aabbb4lf:4,a:1
[5/5]aabba3lf:3
[5/6]bbbba2lf:2,rf:1,rp:1,a:3
[6/0]aaabbb8lf:4,lp:4,a:1
[6/1]aaabba6lf:3,lp:3
[6/2]bbbbaa4lf:2,ls:2,rf:1,rp:1,rs:1,re:1,lsrp:1,lsrs:1,a:3
[6/3]bbabaa4lf:2,ls:2,rf:1,rp:1,rs:1,re:1,lsrp:1,lsrs:1,a:2
[6/4]babbaa2lf:1,ls:1,rf:1,rp:1,rs:1,re:1,lsrp:1,lsrs:1,a:4
[6/5]abbbaa2lf:1,ls:1,a:1
[6/6]aabbbb2lf:2,a:3
[7/0]aaabbbb4lf:2,lp:2,a:3
[7/1]aaabbba4lf:2,lp:2,a:1
[7/2]aaabbab4lf:2,lp:2,a:2
[7/3]bbbbbaa2lf:1,ls:1,a:6
[7/4]bbbabaa2lf:1,ls:1,a:5
[7/5]bbabbaa2lf:1,ls:1,a:4
[7/6]abbbbaa2lf:1,ls:1,a:3

Considering [length 5 / frequency 0] aaabb=c.

Step 2

Rewriting system is complete. See ⟨a, b | aaabbaa=baa⟩.