Morphocompletion for #2857 ⟨a, b | aaaababaaba=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. abaaaaab ⇒ baabaaaa
2. aaababa ⇒ baaaaab
3. aaaabab ⇒ baabaaa
4. babaabaaaaa ⇒ 1
5. ababaab ⇒ babaaba
6. aaabbaabaaaa ⇒ baaaaabaaaab
7. aaabbabaaba ⇒ baaaaabbaab
8. aaababbaabaaa ⇒ baaaaabaaabab
9. babaabaabaabaaa ⇒ abab
10. babaabaaabaabaaa ⇒ aabab
11. babaabaaaabaabaaa ⇒ aaabab
12. ababbabaabaa ⇒ babaabaabaab
...

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] aba, [3/1] aaa, [3/2] baa, [3/3] aab, [3/4] bab, [3/5] abb, [3/6] bba
Length 4:[4/0] abaa, [4/1] aaba, [4/2] baab, [4/3] baba, [4/4] aaab, [4/5] baaa, [4/6] aaaa
Length 5:[5/0] aabaa, [5/1] abaaa, [5/2] baaba, [5/3] abaab, [5/4] babaa, [5/5] aaaba, [5/6] aaaab
Length 6:[6/0] babaab, [6/1] baabaa, [6/2] aabaaa, [6/3] abaaba, [6/4] aaabab, [6/5] abaaaa, [6/6] aaabba
Length 7:[7/0] baabaaa, [7/1] babaaba, [7/2] abaabaa, [7/3] aabaaba, [7/4] aabaaaa, [7/5] abaaaaa, [7/6] ababbab

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

Step 2

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