Morphocompletion for #4353 ⟨a, b | abbbaaaab=ba⟩

Solved by morph:5/2,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. abbbaaaab ⇒ ba
2. babbaaaab ⇒ abbbaaaba
3. abbbaaaaabbbaaaba ⇒ baabbaaaab
4. baabbaaaabbbbaaaab ⇒ abbbaaaaabbbaaabba
5. babbaaaaabbbaaaba ⇒ abbbaaabaabbaaaab
6. abbbaaaaabbbaaaaabbbaaabba ⇒ baaabbaaaabbbbaaaab
7. baaabbaaaabbbbaaaabbbbaaaab ⇒ abbbaaaaabbbaaaaabbbaaabbba
8. baabbaaaabbbbaaaaabbbaaaba ⇒ abbbaaaaabbbaaabbaabbaaaab
9. babbaaaaabbbaaaaabbbaaabba ⇒ abbbaaabaaabbaaaabbbbaaaab
10. baaaabbaaaabbbbaaaabbbbaaaabaaab ⇒ abbbaaaaabbbaaaaabbbaaaaabbbaaba
11. abbbaaaaabbbaaaaabbbaaaaabbbaaabbba ⇒ baaaabbaaaabbbbaaaabbbbaaaab
...

Collecting factors up to length 7 / frequency 4:

[2/0]aa94lf:94,rf:81
[2/1]bb57lf:57,rf:50
[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]aaa62lf:62,rf:51
[3/1]aab37lf:32,ls:5,rf:30,rs:6
[3/2]baa36lf:32,lp:4,rf:30,rp:3
[3/3]abb34lf:30,lp:4,rf:27,rp:7
[4/0]aaab35lf:30,ls:5,rf:26,rs:6
[4/1]aaaa32lf:32,rf:25
[4/2]baaa32lf:30,lp:2,rf:26,rp:2
[4/3]bbaa27lf:27,rf:25
[5/0]bbaaa27lf:27,rf:23
[5/1]aaaab26lf:22,ls:4,rf:18,rs:6
[5/2]baaaa23lf:22,lp:1,rf:18,rp:1
[5/3]aaabb21lf:21,rf:17,a:2
[6/0]bbaaaa21lf:21,rf:17
[6/1]bbbaaa20lf:20,rf:17
[6/2]abbbaa18lf:14,lp:4,rf:14,rp:7
[6/3]baaaab17lf:12,lp:1,ls:4,rf:11,rp:1,rs:6

Considering [length 5 / frequency 2] baaaa=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: wordLimit

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

Collecting factors up to length 6 / frequency 7:

[2/0]ab5lf:4,lp:1,rf:2,rp:2,a:2
[2/1]bb4lf:4,rf:2,a:2
[2/2]bc4lf:4,rf:2,a:2
[2/3]ca3lf:3,a:3
[2/4]cb2lf:1,ls:1,rf:1,rs:1,a:2
[2/5]cc0rf:1,rs:1,a:5
[3/0]abb5lf:4,lp:1,rf:2,rp:2,a:2
[3/1]bbc4lf:4,rf:2,a:2
[3/2]bca3lf:3,a:3
[3/3]cab3lf:3,a:3
[3/4]bcb2lf:1,ls:1,rf:1,rs:1,a:2
[3/5]bcc0rf:1,rs:1,a:5
[4/0]abbc5lf:4,lp:1,rf:2,rp:2,a:2
[4/1]bbca3lf:3,a:3
[4/2]bcab3lf:3,a:3
[4/3]cabb3lf:3,a:3
[4/4]bbcb2lf:1,ls:1,rf:1,rs:1,a:2
[4/5]bbcc0rf:1,rs:1,a:5
[5/0]abbca4lf:3,lp:1,a:3
[5/1]bbcab3lf:3,a:3
[5/2]bcabb3lf:3,a:3
[5/3]cabbc3lf:3,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 | abbbaaaab=ba⟩.