Morphocompletion for #2005 ⟨a, b | aabbbaba=ab

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabbbaba ⇒ ab
2. ababbbaba ⇒ abb
3. aabbbabb ⇒ abbbbaba
4. abbabbbaba ⇒ abbb
5. ababbbabb ⇒ abbbbbaba
6. abbbbabababa ⇒ aabbbb
7. abbbabbbaba ⇒ abbbb
8. abbabbbabb ⇒ abbbbbbaba
9. abbbbbabababa ⇒ ababbbb
10. abbbbabbbaba ⇒ abbbbb
11. abbbabbbabb ⇒ abbbbbbbaba
12. aabbbbbbbaba ⇒ abbbbabababb
13. abbbbbbabababa ⇒ abbabbbb
14. abbbbbabbbaba ⇒ abbbbbb
15. abbbbabbbabb ⇒ abbbbbbbbaba
16. abbbbbbabbbaba ⇒ abbbbbbb
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] bb, [2/1] ab, [2/2] ba, [2/3] aa
Length 3:[3/0] abb, [3/1] bbb, [3/2] bab, [3/3] aba, [3/4] bba, [3/5] aab
Length 4:[4/0] abbb, [4/1] baba, [4/2] bbab, [4/3] bbba, [4/4] babb, [4/5] bbbb, [4/6] abab
Length 5:[5/0] bbbab, [5/1] bbaba, [5/2] bbabb, [5/3] abbba, [5/4] abbbb, [5/5] babbb, [5/6] bbbbb
Length 6:[6/0] bbbaba, [6/1] bbbabb, [6/2] abbbab, [6/3] babbba, [6/4] abbbbb, [6/5] bababa, [6/6] bbabbb
Length 7:[7/0] abbbabb, [7/1] abbbaba, [7/2] babbbab, [7/3] bbabbba, [7/4] bbbabbb, [7/5] abbbbab, [7/6] abababa

Considering [length 4 / frequency 0] abbb=c.

Step 2

Rewriting system is complete. See a, b | aabbbaba=ab.