Morphocompletion for #1082 ⟨a, b | aabbaa=aab⟩

Back:⟨a, b | aabbaa=aab⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabbaa ⇒ aab
2. aabbbaa ⇒ aabb
3. aababbaa ⇒ aabab
4. aabbbbaa ⇒ aabbb
5. aababbbaa ⇒ aababb
6. aabbabbaa ⇒ aabbab
7. aabbbbbaa ⇒ aabbbb
...

Collecting factors up to length 8 / frequency 7:

[2/0]aa28lf:14,lp:7,ls:7,rf:7,rp:7,lprp:7,lsrp:7
[2/1]bb15lf:15,rf:8,rs:4
[2/2]ba10lf:10,rf:3
[2/3]ab10lf:10,rf:10,rs:3
[3/0]baa14lf:7,ls:7
[3/1]aab14lf:7,lp:7,rf:7,rp:7,rs:1,re:1,lprp:7,lprs:1
[3/2]bba8lf:8,rf:1
[3/3]abb8lf:8,rf:5,rs:2
[3/4]bbb7lf:7,rf:3,rs:2,a:1
[3/5]bab3lf:3,rf:3,rs:2,a:2
[3/6]aba2lf:2,rf:2,a:2
[4/0]bbaa14lf:7,ls:7
[4/1]aabb10lf:5,lp:5,rf:4,rp:4,rs:1,re:1,lprp:4,lprs:1
[4/2]bbba4lf:4,a:1
[4/3]abbb4lf:4,rf:2,rs:1,a:1
[4/4]abba4lf:4,rf:1
[4/5]aaba4lf:2,lp:2,rf:2,rp:2,lprp:2,a:2
[4/6]bbbb3lf:3,rf:1,rs:1,a:3
[5/0]bbbaa8lf:4,ls:4,a:1
[5/1]abbaa6lf:3,ls:3
[5/2]aabbb6lf:3,lp:3,rf:2,rp:2,rs:1,re:1,lprp:2,lprs:1,a:1
[5/3]aabba4lf:2,lp:2,rf:1,rp:1,lprp:1
[5/4]aabab4lf:2,lp:2,rf:2,rp:2,rs:1,re:1,lprp:2,lprs:1,a:2
[5/5]bbbba2lf:2,a:3
[5/6]babba2lf:2,a:2
[6/0]bbbbaa4lf:2,ls:2,a:3
[6/1]babbaa4lf:2,ls:2,a:2
[6/2]abbbaa4lf:2,ls:2,a:1
[6/3]aabbbb4lf:2,lp:2,rf:1,rp:1,rs:1,re:1,lprp:1,lprs:1,a:3
[6/4]aababb4lf:2,lp:2,rf:1,rp:1,rs:1,re:1,lprp:1,lprs:1,a:2
[6/5]aabbba2lf:1,lp:1,a:1
[6/6]aabbab2lf:1,lp:1,rf:1,rp:1,rs:1,re:1,lprp:1,lprs:1,a:5
[7/0]bbbbbaa2lf:1,ls:1,a:6
[7/1]bbabbaa2lf:1,ls:1,a:5
[7/2]babbbaa2lf:1,ls:1,a:4
[7/3]abbbbaa2lf:1,ls:1,a:3
[7/4]ababbaa2lf:1,ls:1,a:2
[7/5]aabbbbb2lf:1,lp:1,a:6
[7/6]aabbbba2lf:1,lp:1,a:3

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

Step 2

Rewriting system is complete. See ⟨a, b | aabbaa=aab⟩.