Morphocompletion for #1465 ⟨a, b | aabbbabbba=1⟩

Solved by morph:4/0,2/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. bbbaa ⇒ aabbb
2. baaabbba ⇒ aaabbbab
3. bbbabbba ⇒ abbbabbb
4. baaaaabbb ⇒ aaabbbaba
5. aaabbbabbb ⇒ 1
6. bbbaabbbabbb ⇒ abbbabbbbbba
7. aaabbbabaabbb ⇒ baa
8. abbbabbbbbbaa ⇒ bbb
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] bb, [2/1] aa, [2/2] ba, [2/3] ab
Length 3:[3/0] bbb, [3/1] bba, [3/2] abb, [3/3] baa
Length 4:[4/0] abbb, [4/1] bbba, [4/2] aabb, [4/3] aaab
Length 5:[5/0] abbba, [5/1] aabbb, [5/2] bbbab, [5/3] babbb
Length 6:[6/0] aaabbb, [6/1] bbabbb, [6/2] bbbabb, [6/3] abbbab

Considering [length 4 / frequency 0] abbb=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. cca ⇒ acc
2. aacc ⇒ 1
3. caa ⇒ aac
4. baccc ⇒ acccb
5. cbca ⇒ abcc
6. aacacc ⇒ ca
7. bacacc ⇒ acccba
8. cbaa ⇒ abac
9. cbbca ⇒ abbcc
10. bbb ⇒ accc
11. aacacacc ⇒ caca
12. aacabcc ⇒ bca
13. baaca ⇒ aacab
14. cbbaa ⇒ abbac
15. aacabac ⇒ baa
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] aa, [2/1] ca, [2/2] cc, [2/3] ac, [2/4] ba, [2/5] cb, [2/6] bb
Length 3:[3/0] aac, [3/1] acc, [3/2] aca, [3/3] baa, [3/4] bac, [3/5] cbb, [3/6] bca
Length 4:[4/0] aaca, [4/1] cacc, [4/2] acac, [4/3] bbaa, [4/4] cbba, [4/5] bbca, [4/6] baca
Length 5:[5/0] acacc, [5/1] aacab, [5/2] aacac, [5/3] bacac, [5/4] cabac, [5/5] cabcc, [5/6] acaba

Considering [length 2 / frequency 0] aa=d.

Step 3

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