Morphocompletion for #1480 ⟨a, b | abaaaababa=1⟩

Back:⟨a, b | abaaaababa=1⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abaaaab ⇒ baabaaa
2. aababa ⇒ baaaab
3. aaabab ⇒ baabaa
4. babaabaaaa ⇒ 1
5. ababaab ⇒ babaaba
6. ababaabaa ⇒ babaabaaa
7. ababaabba ⇒ babbaaaab
8. babaabaabaabaa ⇒ abab
9. ababaabbaabaa ⇒ bab
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] ab, [2/1] ba, [2/2] aa, [2/3] bb
Length 3:[3/0] aba, [3/1] baa, [3/2] aab, [3/3] bab, [3/4] aaa, [3/5] bba, [3/6] abb
Length 4:[4/0] abaa, [4/1] abab, [4/2] baab, [4/3] baba, [4/4] aaba, [4/5] aaab, [4/6] abba
Length 5:[5/0] ababa, [5/1] abaab, [5/2] aabaa, [5/3] babaa, [5/4] baaba, [5/5] aabab, [5/6] aabba
Length 6:[6/0] babaab, [6/1] baabaa, [6/2] ababaa, [6/3] abaaba, [6/4] abaaaa, [6/5] baabba, [6/6] abaabb
Length 7:[7/0] abaabaa, [7/1] ababaab, [7/2] babaaba, [7/3] abaabba, [7/4] babaabb, [7/5] baabaab, [7/6] bbaabaa

Considering [length 3 / frequency 0] aba=c.

Step 2

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