Morphocompletion for #2087 ⟨a, b | abbbaaab=ba⟩

Solved by morph:4/3,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. abbbaaab ⇒ ba
2. babbaaab ⇒ abbbaaba
3. abbbaaaabbbaaba ⇒ baabbaaab
4. baabbaaabbbbaaab ⇒ abbbaaaabbbaabba
5. babbaaaabbbaaba ⇒ abbbaabaabbaaab
6. abbbaaaabbbaaaabbbaabba ⇒ baaabbaaabbbbaaab
7. baaabbaaabbbbaaabbbbaaab ⇒ abbbaaaabbbaaaabbbaabbba
8. baabbaaabbbbaaaabbbaaba ⇒ abbbaaaabbbaabbaabbaaab
9. babaaabbbbaaabbbbaaabaab ⇒ abbabbbaaaabbbaaaabbbaba
10. babbaaaabbbaaaabbbaabba ⇒ abbbaabaaabbaaabbbbaaab
11. abbabbbaaaabbbaaaabbbaabbba ⇒ babaaabbbbaaabbbbaaab
...

Collecting factors up to length 7 / frequency 4:

[2/0]aa60lf:60,rf:51
[2/1]bb55lf:55,rf:48
[2/2]ba54lf:41,lp:7,ls:6,rf:35,rp:4,rs:5,re:1,lsrp:3,lprs:4
[2/3]ab48lf:39,lp:4,ls:5,rf:37,rp:7,rs:6,lsrp:4,lprs:3
[3/0]aab35lf:30,ls:5,rf:27,rs:6
[3/1]baa33lf:30,lp:3,rf:27,rp:2
[3/2]abb33lf:29,lp:4,rf:26,rp:7
[3/3]bba32lf:29,ls:3,rf:26,rs:2
[4/0]aaab25lf:21,ls:4,rf:18,rs:6
[4/1]bbaa25lf:25,rf:22
[4/2]abbb23lf:20,lp:3,rf:18,rp:6
[4/3]baaa22lf:21,lp:1,rf:18,rp:1
[5/0]bbaaa19lf:19,rf:15
[5/1]bbbaa19lf:19,rf:16
[5/2]abbba18lf:14,lp:3,ls:1,rf:14,rp:6,rs:1
[5/3]baaab17lf:12,lp:1,ls:4,rf:12,rp:1,rs:6
[6/0]abbbaa16lf:13,lp:3,rf:12,rp:6
[6/1]aaabbb15lf:15,rf:10,a:2
[6/2]bbaaab14lf:10,ls:4,rf:9,rs:6
[6/3]bbbaaa13lf:13,rf:10

Considering [length 4 / frequency 3] baaa=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: lhsLimit

#Rule
1. ca ⇒ abbcc
2. ba ⇒ abbcb
3. abbcabbcabbcb ⇒ c
...

Collecting factors up to length 6 / frequency 7:

[2/0]ab4lf:3,lp:1,rf:2,rp:2,a:2
[2/1]bb3lf:3,rf:2,a:2
[2/2]bc3lf:3,rf:2,a:2
[2/3]ca2lf:2,a:3
[2/4]cb2lf:1,ls:1,rf:1,rs:1,a:2
[2/5]cc0rf:1,rs:1,a:5
[3/0]abb4lf:3,lp:1,rf:2,rp:2,a:2
[3/1]bbc3lf:3,rf:2,a:2
[3/2]bca2lf:2,a:3
[3/3]cab2lf:2,a:3
[3/4]bcb2lf:1,ls:1,rf:1,rs:1,a:2
[3/5]bcc0rf:1,rs:1,a:5
[4/0]abbc4lf:3,lp:1,rf:2,rp:2,a:2
[4/1]bbca2lf:2,a:3
[4/2]bcab2lf:2,a:3
[4/3]cabb2lf:2,a:3
[4/4]bbcb2lf:1,ls:1,rf:1,rs:1,a:2
[4/5]bbcc0rf:1,rs:1,a:5
[5/0]abbca3lf:2,lp:1,a:3
[5/1]bbcab2lf:2,a:3
[5/2]bcabb2lf:2,a:3
[5/3]cabbc2lf:2,a:3
[5/4]abbcb2lf:1,ls:1,rf:1,rp:1,rs:1,re:1,a:2
[5/5]abbcc0rf:1,rp:1,rs:1,re:1,a:5

Considering [length 3 / frequency 0] abb=d.

Step 3

Rewriting system is complete. See ⟨a, b | abbbaaab=ba⟩.