Morphocompletion for #315 ⟨a, b | aabbabba=1⟩

Solved by morph:3/0,2/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. bbaa ⇒ aabb
2. bbabba ⇒ abbabb
3. baaabba ⇒ aaabbab
4. aaabbabb ⇒ 1
5. baaaaabb ⇒ aaabbaba
6. bbabbbba ⇒ abbbbabb
7. abbabbbba ⇒ aabbbbabb
8. bbaabbabb ⇒ abbabbbba
9. abbabbbbaa ⇒ bb
10. aaaabbbbabb ⇒ bba
11. aaabbabaabb ⇒ baa
12. aaaabbbbbbabb ⇒ bbbba
...

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] abb, [3/1] bba, [3/2] aaa, [3/3] bbb
Length 4:[4/0] babb, [4/1] bbab, [4/2] abba, [4/3] aabb
Length 5:[5/0] bbabb, [5/1] aaabb, [5/2] bbbba, [5/3] abbab
Length 6:[6/0] abbabb, [6/1] abbbba, [6/2] aaabba, [6/3] aaaabb

Considering [length 3 / frequency 0] abb=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. bb ⇒ accc
7. aacacc ⇒ ca
8. bacacc ⇒ acccba
9. cbaa ⇒ abac
10. aacacacc ⇒ caca
11. aacabcc ⇒ bca
12. baaca ⇒ aacab
13. aacacabcc ⇒ cabca
14. aacabac ⇒ baa
15. baaaac ⇒ aacaba
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] aa, [2/1] ac, [2/2] cc, [2/3] ca, [2/4] ba, [2/5] cb, [2/6] ab
Length 3:[3/0] aac, [3/1] aca, [3/2] acc, [3/3] baa, [3/4] bac, [3/5] cac, [3/6] bcc
Length 4:[4/0] aaca, [4/1] cacc, [4/2] acac, [4/3] abcc, [4/4] acab, [4/5] baaa, [4/6] baca
Length 5:[5/0] aacac, [5/1] acacc, [5/2] aacab, [5/3] cabcc, [5/4] baaaa, [5/5] aaaac, [5/6] bacac

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

Step 3

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