Morphocompletion for #3182 ⟨a, b | abaaaabbaba=1⟩

Solved by morph:6/0,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abaaaabba ⇒ baabaaaab
2. aabbaba ⇒ baaaabb
3. babaabaa ⇒ aaaabbab
4. ababaabaa ⇒ aaaaabbab
5. babaabaaa ⇒ aabaaaabb
6. abaabaaaabb ⇒ 1
7. babaabaaaab ⇒ 1
8. aaaabbabbaaaabb ⇒ baba
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] aa, [2/1] ba, [2/2] ab, [2/3] bb
Length 3:[3/0] aba, [3/1] baa, [3/2] aab, [3/3] aaa
Length 4:[4/0] abaa, [4/1] baba, [4/2] aabb, [4/3] baaa
Length 5:[5/0] babaa, [5/1] aabaa, [5/2] aaaab, [5/3] abaab
Length 6:[6/0] babaab, [6/1] baabaa, [6/2] aaaabb, [6/3] abaaba

Considering [length 6 / frequency 0] babaab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. bca ⇒ abc
2. aabca ⇒ aaabc
3. aaabca ⇒ 1
4. aaaabcc ⇒ c
5. aabcac ⇒ aaabcc
6. aabcaac ⇒ c
7. caaaabc ⇒ c
8. caaaaabc ⇒ ca
9. babc ⇒ ccaac
10. baabc ⇒ ccaaca
11. baaaabc ⇒ b
12. abaaaabc ⇒ ab
13. baaaaabc ⇒ ba
14. cabaab ⇒ babaac
15. babaab ⇒ c
16. baccaaca ⇒ cc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] aa, [2/1] bc, [2/2] ab, [2/3] ca, [2/4] ba, [2/5] ac, [2/6] cc
Length 3:[3/0] abc, [3/1] aab, [3/2] aaa, [3/3] baa, [3/4] bca, [3/5] caa, [3/6] bab
Length 4:[4/0] aabc, [4/1] aaaa, [4/2] aaab, [4/3] baab, [4/4] abca, [4/5] baaa, [4/6] abaa
Length 5:[5/0] aaabc, [5/1] aaaab, [5/2] aabca, [5/3] baaaa, [5/4] abaab, [5/5] caaaa, [5/6] bacca

Considering [length 3 / frequency 0] abc=d.

Step 3

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