Morphocompletion for #2287 ⟨a, b | aabbbba=baa

Solved by morph:5/2,4/0,2/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabbbba ⇒ baa
2. aabbbbbaa ⇒ babaa
3. aabbbbbbaa ⇒ babbaa
4. aabbbbbabaa ⇒ bababaa
5. aabbbbbbbaa ⇒ babbbaa
6. aabbbbbabbaa ⇒ bababbaa
7. aabbbbbbabaa ⇒ babbabaa
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] bb, [2/1] aa, [2/2] ba
Length 3:[3/0] bbb, [3/1] aab, [3/2] baa
Length 4:[4/0] bbbb, [4/1] aabb, [4/2] bbba
Length 5:[5/0] aabbb, [5/1] bbbbb, [5/2] bbbba
Length 6:[6/0] aabbbb, [6/1] bbbbba, [6/2] abbbbb

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. baa ⇒ aac
2. bca ⇒ cac
3. babaca ⇒ aabcca
4. aacccc ⇒ ca
5. babacca ⇒ aabccca
6. bcbaca ⇒ cabcca
7. babbaca ⇒ aabbcca
8. bbbba ⇒ c
9. aacacccc ⇒ baca
10. babaccca ⇒ aabcccca
11. bcbacca ⇒ cabccca
12. bababcca ⇒ aabcbcca
13. babbacca ⇒ aabbccca
14. bcbbaca ⇒ cabbcca
15. aaccacccc ⇒ bbaca
16. cacacccc ⇒ bcca
17. bcbaccca ⇒ cabcccca
18. aacccbcca ⇒ cbaca
19. bcbabcca ⇒ cabcbcca
20. caccacccc ⇒ bbcca
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] cc, [2/1] ca, [2/2] ba, [2/3] ac, [2/4] bc
Length 3:[3/0] ccc, [3/1] cca, [3/2] bab, [3/3] acc, [3/4] bcb
Length 4:[4/0] cccc, [4/1] bcba, [4/2] accc, [4/3] baba, [4/4] baca
Length 5:[5/0] acccc, [5/1] bcbac, [5/2] babac, [5/3] bacca, [5/4] caccc

Considering [length 4 / frequency 0] cccc=d.

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ca ⇒ aad
2. baa ⇒ aac
3. cda ⇒ daad
4. dc ⇒ cd
5. cdda ⇒ ddaad
6. babada ⇒ aabdaad
7. aadadadad ⇒ da
8. cddda ⇒ dddaad
9. babadda ⇒ aabddaad
10. bcbada ⇒ aadbdaad
11. cccc ⇒ d
12. bbbba ⇒ c
13. aadaaddadad ⇒ bda
14. aaaaddadadad ⇒ bada
15. cdddda ⇒ ddddaad
16. bcbadda ⇒ aadbddaad
17. aadadadacd ⇒ dac
18. cddddda ⇒ dddddaad
19. aacccbdaad ⇒ cbada
20. aadadadaccd ⇒ dacc
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] da, [2/1] ad, [2/2] aa, [2/3] cd, [2/4] dd
Length 3:[3/0] ada, [3/1] dad, [3/2] dda, [3/3] aad, [3/4] cdd
Length 4:[4/0] adad, [4/1] dada, [4/2] aada, [4/3] cddd, [4/4] ddda
Length 5:[5/0] dadad, [5/1] adada, [5/2] aadad, [5/3] cdddd, [5/4] bcbad

Considering [length 2 / frequency 0] da=e.

Step 4

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