Morphocompletion for #3271 ⟨a, b | abbaaaabaab=1⟩

Solved by morph:5/1,4/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. baaaabaaba ⇒ aaaabaabab
2. abbaaaaba ⇒ baaaabaab
3. ababbaa ⇒ bbaaaab
4. aaaabaababb ⇒ 1
5. bbaaaabaab ⇒ aaabaababb
6. babbaaaab ⇒ aabaababb
7. baababba ⇒ abaababb
8. baaaabaabbbaa ⇒ abbaaabbaaaab
9. bbaaaabbabba ⇒ abababaababb
10. aaaabaabababaababb ⇒ aababba
11. bbaaaabaaabaababb ⇒ ababba
...

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] baa, [3/1] aaa, [3/2] aba, [3/3] aab
Length 4:[4/0] aaba, [4/1] bbaa, [4/2] babb, [4/3] aaab
Length 5:[5/0] aaaab, [5/1] ababb, [5/2] baaaa, [5/3] aaaba
Length 6:[6/0] baaaab, [6/1] aaaaba, [6/2] bbaaaa, [6/3] aababb

Considering [length 5 / frequency 1] ababb=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. baca ⇒ abac
2. abaca ⇒ aabac
3. aabacaa ⇒ 1
4. abacaaa ⇒ 1
5. caaaab ⇒ abacaa
6. acaaaab ⇒ 1
7. caaaaba ⇒ 1
8. bb ⇒ caaac
9. abb ⇒ acaaac
10. abacaaac ⇒ c
11. caaaaabac ⇒ ca
12. abaabacaa ⇒ ab
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] aa, [2/1] ab, [2/2] ca, [2/3] ac, [2/4] ba, [2/5] bb
Length 3:[3/0] aba, [3/1] caa, [3/2] aaa, [3/3] aca, [3/4] bac, [3/5] aab, [3/6] aac
Length 4:[4/0] abac, [4/1] caaa, [4/2] acaa, [4/3] baca, [4/4] aaba, [4/5] aaab, [4/6] aaaa
Length 5:[5/0] caaaa, [5/1] bacaa, [5/2] abaca, [5/3] aaaab, [5/4] aabac, [5/5] acaaa, [5/6] aaaba

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

Step 3

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