Morphocompletion for #712 ⟨a, b | ababaabba=1⟩

Back:⟨a, b | ababaabba=1⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. baaa ⇒ aaab
2. baaba ⇒ abaab
3. abaabba ⇒ aabbaab
4. aabaabba ⇒ aaabbaab
5. baabbaa ⇒ ababaab
6. aababaabb ⇒ 1
7. aabaabbba ⇒ aaabbbaab
8. babaabba ⇒ ababaabb
9. abaabbbaa ⇒ aabbabaab
10. aaabbabaabb ⇒ ba
11. aaabababaabb ⇒ baa
12. aaabbbabaabb ⇒ bba
13. aaababbabaabb ⇒ baba
14. aaabbababaabb ⇒ bbaa
15. aabababaabbba ⇒ baab
16. aaabbbbabaabb ⇒ bbba
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] aa, [2/1] ba, [2/2] ab, [2/3] bb
Length 3:[3/0] aab, [3/1] abb, [3/2] aba, [3/3] baa, [3/4] bba, [3/5] bab, [3/6] aaa
Length 4:[4/0] aabb, [4/1] baab, [4/2] abaa, [4/3] baba, [4/4] aaba, [4/5] aaab, [4/6] abba
Length 5:[5/0] baabb, [5/1] abaab, [5/2] babaa, [5/3] aabba, [5/4] aaabb, [5/5] abbba, [5/6] ababa
Length 6:[6/0] abaabb, [6/1] babaab, [6/2] baabba, [6/3] aabbba, [6/4] aababa, [6/5] bbabaa, [6/6] aaabbb
Length 7:[7/0] babaabb, [7/1] baabbba, [7/2] bbabaab, [7/3] abaabbb, [7/4] abaabba, [7/5] ababaab, [7/6] aaabbab

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

Step 2

Rewriting system is complete. See ⟨a, b | ababaabba=1⟩.