Morphocompletion for #267 ⟨a, b | abba=bab⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abba ⇒ bab
2. babbba ⇒ abbbab
3. aabbbabb ⇒ babbbba
4. babbbbaa ⇒ aabbbbab
5. babbbbaba ⇒ ababbbbab
6. abaabbbbab ⇒ babbbbbaa
7. abbbabbbba ⇒ bbabbbbab
8. abababbbbab ⇒ babbbbbaba
9. abbbabbbbba ⇒ bbbabbbbab
10. babbbbbaaba ⇒ abaabbbbbab
11. abaababbbbab ⇒ babbbbbabaa
...

Collecting factors up to length 7 / frequency 4:

[2/0]bb37lf:36,ls:1,rf:36,rp:2
[2/1]ba34lf:24,lp:4,ls:6,rf:21,rp:5,rs:2,lsrp:1
[2/2]ab33lf:24,lp:6,ls:3,rf:21,rp:3,rs:7,lprs:3
[2/3]aa7lf:5,lp:1,ls:1,rf:4,rp:1,rs:2,lsrp:1,a:2
[3/0]bbb22lf:22,rf:24,rp:1,a:1
[3/1]bab21lf:14,lp:4,ls:3,rf:16,rp:5,rs:7,re:1,lsrp:3,lprs:4
[3/2]abb18lf:14,lp:3,ls:1,rf:10,rp:1
[3/3]bba17lf:13,ls:4,rf:12,rp:1,rs:1,lsrp:1
[4/0]bbba15lf:12,ls:3,rf:11,rp:1,rs:1,lsrp:1,a:1
[4/1]babb14lf:9,lp:4,ls:1,rf:7,rp:4,lsrp:1,a:1
[4/2]abbb14lf:12,lp:2,rf:10,rp:1,a:1
[4/3]bbbb10lf:10,rf:13,a:2
[5/0]babbb12lf:8,lp:4,rf:7,rp:4,a:1
[5/1]bbbba10lf:8,ls:2,rf:9,rs:1,a:2
[5/2]bbbab10lf:7,ls:3,rf:9,rp:1,rs:6,a:1
[5/3]abbbb8lf:8,rf:9,a:2
[6/0]babbbb10lf:7,lp:3,rf:7,rp:4,a:2
[6/1]bbbbab7lf:4,ls:3,rf:7,rs:5,a:3
[6/2]abbbba7lf:6,ls:1,rf:5,rs:1,a:2
[6/3]abbbab5lf:3,lp:2,rf:1,rp:1,rs:1,re:1,a:1

Considering [length 2 / frequency 1] ba=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ba ⇒ c
2. abc ⇒ cb
3. bcb ⇒ cbc
4. cbca ⇒ bcc
5. acbc ⇒ cbb
6. cbcca ⇒ bbcc
7. accbc ⇒ cbbb
8. abbcc ⇒ cbbca
9. bccbc ⇒ cbccb
10. cbccca ⇒ bbbcc
11. acccbc ⇒ cbbbb
12. acbbcc ⇒ cbbbca
13. abbbcc ⇒ cbbcca
14. bcccbc ⇒ cbccbb
15. cbcccca ⇒ bbbbcc
16. accccbc ⇒ cbbbbb
17. accbbcc ⇒ cbbbbca
18. acbbbcc ⇒ cbbbcca
19. abbbbcc ⇒ cbbccca
20. bccccbc ⇒ cbccbbb
...

Collecting factors up to length 6 / frequency 7:

[2/0]bc34lf:22,lp:4,ls:8,rf:14,rp:1,rs:1,lprs:10,a:2
[2/1]cc31lf:25,ls:6,rf:11,rs:4,a:5
[2/2]cb20lf:15,lp:4,ls:1,rf:18,rp:15,rs:2,re:1,lsrp:10,a:2
[2/3]ac14lf:7,lp:7,a:4
[2/4]bb10lf:10,rf:29,rp:3,rs:6,a:4
[2/5]ab8lf:4,lp:4,a:2
[2/6]ca8lf:4,ls:4,rf:6,rs:6,a:5
[3/0]cbc22lf:11,lp:4,ls:7,rf:4,rp:4,rs:1,re:1,lsrp:3,a:3
[3/1]bcc21lf:12,lp:3,ls:6,rf:10,rp:1,rs:4,re:1,a:5
[3/2]ccc9lf:9,rf:1,a:10
[3/3]acc8lf:4,lp:4,a:9
[3/4]ccb7lf:7,rf:3,rs:1,a:6
[3/5]bbc6lf:6,rf:9,rp:1,a:7
[3/6]abb6lf:3,lp:3,a:7
[4/0]bbcc12lf:6,ls:6,rf:6,rp:1,rs:3,re:1,a:7
[4/1]ccbc12lf:6,ls:6,a:6
[4/2]bccc6lf:4,lp:2,rf:1,a:10
[4/3]cbcc6lf:3,lp:3,rf:3,rp:3,a:6
[4/4]abbb4lf:2,lp:2,a:11
[4/5]acbb4lf:2,lp:2,a:15
[4/6]accb4lf:2,lp:2,a:9
[5/0]cccbc8lf:4,ls:4,a:10
[5/1]bbbcc6lf:3,ls:3,rf:3,rp:1,rs:2,re:1,a:11
[5/2]cbbcc4lf:2,ls:2,rf:2,rp:2,a:11
[5/3]cbccc4lf:2,lp:2,a:12
[5/4]bcccc3lf:2,lp:1,a:13
[5/5]bcccb2lf:1,lp:1,a:10
[5/6]bccca2lf:1,ls:1,rf:1,rs:1,a:12

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

Step 3

Rewriting system is complete. See ⟨a, b | abba=bab⟩.