Morphocompletion for #4327 ⟨a, b | abbaaaaab=ba⟩

Solved by morph:6/1,2/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. abbaaaaab ⇒ ba
2. babaaaaab ⇒ abbaaaaba
3. abbaaaaaabbaaaaba ⇒ baabaaaaab
4. baabaaaaabbbaaaaab ⇒ abbaaaaaabbaaaabba
5. babaaaaaabbaaaaba ⇒ abbaaaabaabaaaaab
6. baaabaaaaabbbaaaaabaaaab ⇒ abbaaaaaabbaaaaaabbaaaba
7. abbaaaaaabbaaaaaabbaaaabba ⇒ baaabaaaaabbbaaaaab
8. baaabaaaaabbbaaaaabbbaaaaab ⇒ abbaaaaaabbaaaaaabbaaaabbba
9. baabaaaaabbbaaaaaabbaaaaba ⇒ abbaaaaaabbaaaabbaabaaaaab
10. babaaaaaabbaaaaaabbaaaabba ⇒ abbaaaabaaabaaaaabbbaaaaab
11. abbaaaaaabbaaaaaabbaaaaaabbaaaba ⇒ baaaabaaaaabbbaaaaabaaaab
12. abbaaaaaabbaaaaaabbaaaaaabbaaaabbba ⇒ baaaabaaaaabbbaaaaabbbaaaaab
...

Collecting factors up to length 7 / frequency 4:

[2/0]aa134lf:134,rf:114
[2/1]ba59lf:45,lp:7,ls:7,rf:38,rp:5,rs:5,re:1,lsrp:4,lprs:4
[2/2]ab53lf:43,lp:5,ls:5,rf:40,rp:7,rs:7,lsrp:4,lprs:4
[2/3]bb32lf:32,rf:27
[3/0]aaa99lf:99,rf:81
[3/1]aab40lf:35,ls:5,rf:33,rs:7
[3/2]baa39lf:35,lp:4,rf:33,rp:4
[3/3]abb31lf:26,lp:5,rf:21,rp:7
[4/0]aaaa66lf:66,rf:51
[4/1]aaab38lf:33,ls:5,rf:30,rs:7
[4/2]baaa35lf:33,lp:2,rf:30,rp:3
[4/3]abba27lf:20,lp:5,ls:2,rf:15,rp:7,rs:1,lsrp:1
[5/0]aaaaa36lf:36,rf:24
[5/1]aaaab35lf:30,ls:5,rf:27,rs:7
[5/2]baaaa30lf:30,rf:27,rp:2
[5/3]bbaaa23lf:23,rf:18
[6/0]aaaaab27lf:23,ls:4,rf:18,rs:6
[6/1]baaaaa23lf:23,rf:18
[6/2]abbaaa23lf:18,lp:5,rf:13,rp:7
[6/3]bbaaaa22lf:22,rf:17

Considering [length 6 / frequency 1] baaaaa=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: wordLimit

#Rule
1. ca ⇒ abcc
2. ba ⇒ abcb
3. abcabcabcabcabcb ⇒ c
...

Collecting factors up to length 6 / frequency 7:

[2/0]ab6lf:5,lp:1,rf:2,rp:2,a:2
[2/1]bc5lf:5,rf:2,a:2
[2/2]ca4lf:4,a:3
[2/3]cb2lf:1,ls:1,rf:1,rs:1,a:2
[2/4]cc0rf:1,rs:1,a:5
[3/0]abc6lf:5,lp:1,rf:2,rp:2,a:2
[3/1]bca4lf:4,a:3
[3/2]cab4lf:4,a:3
[3/3]bcb2lf:1,ls:1,rf:1,rs:1,a:2
[3/4]bcc0rf:1,rs:1,a:5
[4/0]abca5lf:4,lp:1,a:3
[4/1]bcab4lf:4,a:3
[4/2]cabc4lf:4,a:3
[4/3]abcb2lf:1,ls:1,rf:1,rp:1,rs:1,re:1,a:2
[4/4]abcc0rf:1,rp:1,rs:1,re:1,a:5
[5/0]abcab5lf:4,lp:1,a:3
[5/1]bcabc4lf:4,a:3
[5/2]cabca3lf:3,a:3
[5/3]cabcb2lf:1,ls:1,a:3

Considering [length 2 / frequency 0] ab=d.

Step 3

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