Morphocompletion for #1410 ⟨a, b | aabaabbaba=1⟩

Solved by morph:5/0,4/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabaabba ⇒ baaabaab
2. aabbaba ⇒ babaaab
3. babaaaba ⇒ aaabaabb
4. bbabaa ⇒ abaabb
5. abaaabaabb ⇒ 1
6. babaaabaab ⇒ 1
7. baabbaba ⇒ abaabbab
8. abaaabaababaabb ⇒ babaa
9. abaabbbbaba ⇒ bbbaaabaabb
10. abaabbbaaabaabb ⇒ bbaba
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ba, [2/1] ab, [2/2] aa, [2/3] bb
Length 3:[3/0] aba, [3/1] aab, [3/2] baa, [3/3] abb
Length 4:[4/0] abaa, [4/1] baba, [4/2] aabb, [4/3] baab
Length 5:[5/0] baabb, [5/1] abaab, [5/2] bbaba, [5/3] babaa
Length 6:[6/0] abaabb, [6/1] aabaab, [6/2] baaaba, [6/3] abaaab

Considering [length 5 / frequency 0] baabb=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aaacab ⇒ 1
2. caba ⇒ acab
3. baaac ⇒ aacab
4. acac ⇒ bb
5. aaacaac ⇒ babaa
6. baabb ⇒ c
7. aabbaba ⇒ babaaab
8. babaaaba ⇒ aaac
9. bbabaa ⇒ ac
10. acaacab ⇒ bbaba
11. baabc ⇒ caabb
12. bbac ⇒ acbb
13. baabac ⇒ cbabaa
14. aacabbb ⇒ cac
15. baaacbb ⇒ cac
16. acabcac ⇒ cabbb
17. bbababb ⇒ accac
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ba, [2/1] ac, [2/2] aa, [2/3] ab, [2/4] bb, [2/5] ca, [2/6] bc
Length 3:[3/0] baa, [3/1] aca, [3/2] aac, [3/3] aba, [3/4] cab, [3/5] bba, [3/6] aaa
Length 4:[4/0] acab, [4/1] aaac, [4/2] baba, [4/3] baab, [4/4] bbab, [4/5] aaca, [4/6] baaa
Length 5:[5/0] bbaba, [5/1] aacab, [5/2] acaac, [5/3] babaa, [5/4] aaaca, [5/5] cabbb, [5/6] abcac

Considering [length 4 / frequency 0] acab=d.

Step 3

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