Morphocompletion for #2317 ⟨a, b | abaabba=baa

Solved by morph:5/1,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abaabba ⇒ baa
2. abaabbbaa ⇒ babaa
3. abaabbbbaa ⇒ babbaa
4. abaabbbabaa ⇒ bababaa
5. abaabbbbbaa ⇒ babbbaa
6. abaabbbbabaa ⇒ babbabaa
7. abaabbbabbaa ⇒ bababbaa
8. abaabbbababaa ⇒ babababaa
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ab, [2/1] ba, [2/2] aa, [2/3] bb
Length 3:[3/0] baa, [3/1] aba, [3/2] bbb, [3/3] bba
Length 4:[4/0] abaa, [4/1] aabb, [4/2] baab, [4/3] bbaa
Length 5:[5/0] abaab, [5/1] baabb, [5/2] aabbb, [5/3] bbbaa
Length 6:[6/0] abaabb, [6/1] baabbb, [6/2] bbbbaa, [6/3] aabbba

Considering [length 5 / frequency 1] baabb=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. baa ⇒ aca
2. bac ⇒ acc
3. acaacca ⇒ caa
4. acabcaa ⇒ caaacca
5. caabb ⇒ acabc
6. acabb ⇒ c
7. babc ⇒ acbc
8. accaacca ⇒ bcaa
9. acaaccc ⇒ cac
10. accabb ⇒ bc
11. acabcac ⇒ caaaccc
12. babbc ⇒ acbbc
13. acccaacca ⇒ bbcaa
14. accaaccc ⇒ bcac
15. acaaccbc ⇒ cabc
16. acccabb ⇒ bbc
17. babbbc ⇒ acbbbc
18. acccaaccc ⇒ bbcac
19. accaaccbc ⇒ bcabc
20. accccabb ⇒ bbbc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ac, [2/1] cc, [2/2] ca, [2/3] bb, [2/4] aa, [2/5] bc, [2/6] ab
Length 3:[3/0] acc, [3/1] cca, [3/2] abb, [3/3] aca, [3/4] caa, [3/5] ccc, [3/6] aac
Length 4:[4/0] accc, [4/1] acca, [4/2] cabb, [4/3] aacc, [4/4] caac, [4/5] acab, [4/6] acaa
Length 5:[5/0] caacc, [5/1] ccabb, [5/2] aaccc, [5/3] accca, [5/4] aacca, [5/5] acaac, [5/6] accaa

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

Step 3

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