Morphocompletion for #3210 ⟨a, b | abaabababba=1⟩

Solved by morph:2/0,5/2. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abbaaba ⇒ baabaab
2. babbaa ⇒ ababab
3. aabbaabaab ⇒ abaabaabab
4. abababba ⇒ bbaabaab
5. aabababba ⇒ baabaabab
6. baabaababa ⇒ aabaababab
7. aabaabababb ⇒ 1
8. abaabababababab ⇒ bbaa
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ba, [2/1] ab, [2/2] aa, [2/3] bb
Length 3:[3/0] aba, [3/1] bab, [3/2] aab, [3/3] baa
Length 4:[4/0] abab, [4/1] aaba, [4/2] baba, [4/3] baab
Length 5:[5/0] ababa, [5/1] babab, [5/2] baaba, [5/3] babba
Length 6:[6/0] ababab, [6/1] aababa, [6/2] aabaab, [6/3] bababb

Considering [length 2 / frequency 0] ba=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. cacacca ⇒ aacaccc
2. cacaccc ⇒ a
3. aacaccb ⇒ cacacc
4. cacaccb ⇒ 1
5. aacaccccaccb ⇒ cacac
6. ba ⇒ c
7. aaccbc ⇒ cacacb
8. acaccbc ⇒ 1
9. ccaccbc ⇒ b
10. aacacccccbc ⇒ cacab
11. bca ⇒ acacccccb
12. acaccbb ⇒ caccbc
13. aaccccaccbb ⇒ acbc
14. acaccccaccbb ⇒ accbc
15. ccaccccaccbb ⇒ cccbc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] cc, [2/1] ac, [2/2] ca, [2/3] cb, [2/4] bc, [2/5] aa, [2/6] bb
Length 3:[3/0] cac, [3/1] acc, [3/2] ccb, [3/3] ccc, [3/4] aca, [3/5] cca, [3/6] aac
Length 4:[4/0] cacc, [4/1] accb, [4/2] acac, [4/3] ccbb, [4/4] ccbc, [4/5] ccac, [4/6] accc
Length 5:[5/0] caccb, [5/1] acacc, [5/2] accbb, [5/3] ccacc, [5/4] accbc, [5/5] caccc, [5/6] cacac

Considering [length 5 / frequency 2] accbb=d.

Step 3

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