Morphocompletion for #2265 ⟨a, b | aabbaba=baa

Solved by morph:4/1,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabbaba ⇒ baa
2. aabbabbaa ⇒ babaa
3. aabbabbbaa ⇒ babbaa
4. aabbabbbbaa ⇒ babbbaa
5. aabbabbbbbaa ⇒ babbbbaa
6. aabbabbabaa ⇒ bababaa
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] aa, [2/1] bb, [2/2] ba, [2/3] ab
Length 3:[3/0] aab, [3/1] abb, [3/2] bba, [3/3] baa
Length 4:[4/0] aabb, [4/1] abba, [4/2] bbaa, [4/3] bbab
Length 5:[5/0] aabba, [5/1] abbab, [5/2] bbbaa, [5/3] bbabb
Length 6:[6/0] aabbab, [6/1] abbabb, [6/2] bbbbaa, [6/3] abbaba

Considering [length 4 / frequency 1] abba=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. acbc ⇒ bac
2. abbc ⇒ cbba
3. acbbcc ⇒ babcc
4. abbbcc ⇒ ccbbbac
5. acbbbcc ⇒ bcbbac
6. acba ⇒ baa
7. abba ⇒ c
8. acbbca ⇒ babca
9. abbbca ⇒ ccbbbaa
10. acbbbca ⇒ bcbbaa
11. acbbac ⇒ babac
12. abbbac ⇒ ccbc
13. acbbbac ⇒ bcc
14. abbbbac ⇒ ccbcbc
15. acbbbbac ⇒ bccbc
16. acbbaa ⇒ babaa
17. abbbaa ⇒ ccba
18. acbbbaa ⇒ bca
19. abbbbaa ⇒ ccbcba
20. acbbbbaa ⇒ bccba
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ac, [2/1] bb, [2/2] ab, [2/3] ba, [2/4] cb, [2/5] aa, [2/6] bc
Length 3:[3/0] acb, [3/1] abb, [3/2] bbb, [3/3] bba, [3/4] baa, [3/5] bac, [3/6] cbb
Length 4:[4/0] acbb, [4/1] abbb, [4/2] bbaa, [4/3] bbac, [4/4] bbba, [4/5] bbca, [4/6] cbbb
Length 5:[5/0] acbbb, [5/1] bbbaa, [5/2] bbbac, [5/3] abbba, [5/4] acbba, [5/5] abbbb, [5/6] abbbc

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

Step 3

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