Morphocompletion for #1789 ⟨a, b | ababbaaab=a

Solved by morph:3/2. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aaaabb ⇒ aabbaa
2. abaaabb ⇒ ababbaa
3. aabbaaab ⇒ ababbaaa
4. aaababbaaa ⇒ aabbaaaaab
5. aaaabaabb ⇒ aabbaabaa
6. ababbaaab ⇒ a
7. abaababbaaa ⇒ ababbaaaaab
8. abaaabaabb ⇒ ababbaabaa
9. aaaababaabb ⇒ aaaababbbaa
10. aaaabaabaabb ⇒ aabbaabaabaa
11. abaaababaabb ⇒ abaaababbbaa
12. aabbaababbaaa ⇒ aaaab
13. ababbaabbaaaaab ⇒ aabbaaa
14. ababbaababbaaa ⇒ abaaab
15. aabbaababbaabbaa ⇒ aaaababb
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] aa, [2/1] ab, [2/2] ba, [2/3] bb
Length 3:[3/0] aab, [3/1] aaa, [3/2] abb, [3/3] aba, [3/4] baa, [3/5] bba, [3/6] bab
Length 4:[4/0] aabb, [4/1] aaab, [4/2] bbaa, [4/3] abaa, [4/4] baaa, [4/5] abab, [4/6] abba
Length 5:[5/0] abbaa, [5/1] baabb, [5/2] ababb, [5/3] bbaaa, [5/4] aaaab, [5/5] babba, [5/6] aabba
Length 6:[6/0] ababba, [6/1] abbaaa, [6/2] abaabb, [6/3] aabbaa, [6/4] babbaa, [6/5] abaaab, [6/6] aaaaba
Length 7:[7/0] ababbaa, [7/1] babbaaa, [7/2] aabaabb, [7/3] aababba, [7/4] baababb, [7/5] babaabb, [7/6] aabbaab

Considering [length 3 / frequency 2] abb=c.

Step 2

Rewriting system is complete. See a, b | ababbaaab=a.