Morphocompletion for #4807 ⟨a, b | abaabbba=baa

Solved by morph:6/1,3/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abaabbba ⇒ baa
2. abaabbbbaa ⇒ babaa
3. abaabbbbbaa ⇒ babbaa
4. abaabbbbabaa ⇒ bababaa
5. abaabbbbbbaa ⇒ babbbaa
6. abaabbbbabbaa ⇒ bababbaa
7. abaabbbbbabaa ⇒ babbabaa
8. abaabbbbababaa ⇒ babababaa
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ab, [2/1] bb, [2/2] ba, [2/3] aa
Length 3:[3/0] baa, [3/1] aba, [3/2] bbb, [3/3] bba
Length 4:[4/0] abaa, [4/1] bbbb, [4/2] bbba, [4/3] abbb
Length 5:[5/0] abaab, [5/1] baabb, [5/2] aabbb, [5/3] bbbba
Length 6:[6/0] abaabb, [6/1] baabbb, [6/2] aabbbb, [6/3] bbbbaa

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. acaaccca ⇒ caa
2. acaacccc ⇒ cac
3. acaacccbc ⇒ cabc
4. baa ⇒ aca
5. bac ⇒ acc
6. bcaa ⇒ accaaccca
7. bcac ⇒ accaacccc
8. acaacccbbc ⇒ cabbc
9. babc ⇒ acbc
10. bcabc ⇒ accaacccbc
11. caabbb ⇒ acabbc
12. acabbb ⇒ c
13. accabbb ⇒ bc
14. acccabbb ⇒ bbc
15. accaaccccabbb ⇒ bcc
16. acccaaccccabbb ⇒ bbcc
17. accaacccccabbb ⇒ bcbc
18. bbbc ⇒ accccabbb
19. babbc ⇒ acbbc
20. bcabbc ⇒ accaacccbbc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ac, [2/1] bb, [2/2] cc, [2/3] bc, [2/4] ca, [2/5] aa, [2/6] ab
Length 3:[3/0] acc, [3/1] bbb, [3/2] ccc, [3/3] caa, [3/4] cca, [3/5] aca, [3/6] abb
Length 4:[4/0] abbb, [4/1] accc, [4/2] acaa, [4/3] cabb, [4/4] aacc, [4/5] caac, [4/6] ccca
Length 5:[5/0] cabbb, [5/1] acaac, [5/2] aaccc, [5/3] caacc, [5/4] accca, [5/5] ccabb, [5/6] acccc

Considering [length 3 / frequency 1] bbb=d.

Step 3

Rewriting system is complete. See a, b | abaabbba=baa.