Morphocompletion for #2942 ⟨a, b | aaababbaaba=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. bbaabaa ⇒ aaababb
2. aabaaaababb ⇒ 1
3. aababbaab ⇒ baaaababb
4. baaaababba ⇒ abaaaababb
5. babbaaba ⇒ ababbaab
6. aabaaaababaaababb ⇒ baabaa
7. aaababbbaaaababb ⇒ bbaab
...

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] aab, [3/1] aba, [3/2] baa, [3/3] aaa
Length 4:[4/0] aaba, [4/1] babb, [4/2] abab, [4/3] aaab
Length 5:[5/0] ababb, [5/1] aabab, [5/2] aaaba, [5/3] aabaa
Length 6:[6/0] aababb, [6/1] aaabab, [6/2] baaaab, [6/3] bbaaba

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aaacaab ⇒ 1
2. caaba ⇒ acaab
3. baaac ⇒ acaab
4. aacac ⇒ bb
5. aaacaaaac ⇒ baabaa
6. ababb ⇒ c
7. bbaabaa ⇒ aac
8. aacaacaab ⇒ bbaaba
9. ababaac ⇒ cbaabaa
10. aacaabc ⇒ babb
11. babbaaba ⇒ caab
12. acaabbb ⇒ cac
13. ababacaab ⇒ caaac
14. babbac ⇒ caabbb
15. bbaabac ⇒ aacbabb
16. ababcac ⇒ caaacbb
17. bbaabbb ⇒ aaccac
18. acaabbabb ⇒ caabc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] aa, [2/1] ab, [2/2] ac, [2/3] ba, [2/4] bb, [2/5] ca, [2/6] bc
Length 3:[3/0] aba, [3/1] aac, [3/2] aab, [3/3] aca, [3/4] caa, [3/5] bba, [3/6] bab
Length 4:[4/0] caab, [4/1] acaa, [4/2] aaca, [4/3] babb, [4/4] abab, [4/5] aaac, [4/6] bbaa
Length 5:[5/0] acaab, [5/1] bbaab, [5/2] aacaa, [5/3] babba, [5/4] aabbb, [5/5] baaba, [5/6] ababa

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

Step 3

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