Morphocompletion for #1328 ⟨a, b | aaaabbabba=1⟩

Solved by morph:4/0,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aaaabbaaa ⇒ bbaaaaaaa
2. bbabbaaaaa ⇒ 1
3. abbabb ⇒ bbabba
4. bbabbbbaaaaaaaa ⇒ aaabb
5. bbabbaabbaaaa ⇒ abb
6. bbabbaaabbaaaa ⇒ aabb
7. abbbbabba ⇒ bbabbaabb
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] aa, [2/1] bb, [2/2] ba, [2/3] ab
Length 3:[3/0] aaa, [3/1] bba, [3/2] abb, [3/3] baa
Length 4:[4/0] aaaa, [4/1] bbab, [4/2] abba, [4/3] babb
Length 5:[5/0] bbabb, [5/1] bbaaa, [5/2] aaaaa, [5/3] abbaa
Length 6:[6/0] bbabba, [6/1] abbaaa, [6/2] bbaaaa, [6/3] aaaaaa

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ac ⇒ ca
2. aaaa ⇒ c
3. cbb ⇒ bbc
4. cabb ⇒ abbc
5. caabb ⇒ aabbc
6. caaabb ⇒ aaabbc
7. abbcab ⇒ babbca
8. bbabbcc ⇒ aaa
9. bbabbca ⇒ 1
10. cbbabbc ⇒ aaa
11. cbbabba ⇒ 1
12. abbabb ⇒ bbabba
13. abbabbc ⇒ 1
14. cabbabb ⇒ 1
15. bbcbabbca ⇒ cb
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] bb, [2/1] ab, [2/2] ca, [2/3] ba, [2/4] bc, [2/5] aa, [2/6] cb
Length 3:[3/0] abb, [3/1] bba, [3/2] bbc, [3/3] bab, [3/4] cab, [3/5] bca, [3/6] aaa
Length 4:[4/0] babb, [4/1] bbab, [4/2] abbc, [4/3] abba, [4/4] bbca, [4/5] aabb, [4/6] cbba
Length 5:[5/0] bbabb, [5/1] babbc, [5/2] abbca, [5/3] abbab, [5/4] cbbab, [5/5] babba, [5/6] bbcab

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

Step 3

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