Morphocompletion for #1492 ⟨a, b | abaabaabab=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. abababa ⇒ babaaba
2. aabaab ⇒ abaaba
3. baabaaab ⇒ ababaaba
4. bababaabaa ⇒ 1
5. abababaaba ⇒ 1
6. abaabaaab ⇒ aababaaba
7. abaabaaaab ⇒ aaababaaba
8. babaaababaaba ⇒ aab
9. abaabaaaaab ⇒ aaaababaaba
10. abaabaaaaaab ⇒ aaaaababaaba
11. abaabaaaaaaab ⇒ aaaaaababaaba
12. abaabaaaaaaaab ⇒ aaaaaaababaaba
13. abaabaaaaaaaaab ⇒ aaaaaaaababaaba
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] ab, [2/1] aa, [2/2] ba
Length 3:[3/0] aba, [3/1] aab, [3/2] aaa, [3/3] baa, [3/4] bab
Length 4:[4/0] abaa, [4/1] aaaa, [4/2] aaab, [4/3] aaba, [4/4] baab, [4/5] baba, [4/6] baaa
Length 5:[5/0] abaab, [5/1] aaaaa, [5/2] baaba, [5/3] aaaab, [5/4] aabaa, [5/5] abaaa, [5/6] ababa
Length 6:[6/0] abaaba, [6/1] baabaa, [6/2] aaaaaa, [6/3] aaaaab, [6/4] aabaaa, [6/5] abaaaa, [6/6] baaaaa
Length 7:[7/0] abaabaa, [7/1] baabaaa, [7/2] aaaaaab, [7/3] aaaaaaa, [7/4] aabaaaa, [7/5] abaaaaa, [7/6] babaaba

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

Step 2

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