Morphocompletion for #3285 ⟨a, b | abbabaabaab=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. ababb ⇒ bbaba
2. aababb ⇒ abbaba
3. bababb ⇒ bbbaba
4. aabaabab ⇒ babaabaa
5. bbaababb ⇒ bbabbaba
6. babaabaab ⇒ abbabaaba
7. babaababb ⇒ bbbabaaba
8. babbabaabaa ⇒ 1
9. ababbabaaba ⇒ 1
10. ababbbbabaaba ⇒ bb
11. bbabaabbabaabaa ⇒ abab
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ab, [2/1] ba, [2/2] bb, [2/3] aa
Length 3:[3/0] aba, [3/1] bab, [3/2] aab, [3/3] abb
Length 4:[4/0] babb, [4/1] aaba, [4/2] abaa, [4/3] abab
Length 5:[5/0] ababb, [5/1] abaab, [5/2] baaba, [5/3] babaa
Length 6:[6/0] babaab, [6/1] abaaba, [6/2] bbabaa, [6/3] baabaa

Considering [length 3 / frequency 0] aba=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. bccc ⇒ cccb
2. accb ⇒ bcca
3. acccb ⇒ cbacc
4. abc ⇒ cba
5. cccbb ⇒ 1
6. bbc ⇒ cbb
7. cbbc ⇒ ccbb
8. cbbcc ⇒ 1
9. aba ⇒ c
10. cbaccb ⇒ a
11. cccbcbb ⇒ bc
12. cbaccbb ⇒ ab
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] cb, [2/1] cc, [2/2] bb, [2/3] bc, [2/4] ac, [2/5] ab, [2/6] ba
Length 3:[3/0] cbb, [3/1] ccb, [3/2] ccc, [3/3] acc, [3/4] cba, [3/5] bcc, [3/6] bbc
Length 4:[4/0] cccb, [4/1] ccbb, [4/2] cbac, [4/3] accb, [4/4] bcbb, [4/5] bbcc, [4/6] cbbc
Length 5:[5/0] cbacc, [5/1] baccb, [5/2] cbcbb, [5/3] accbb, [5/4] cccbc, [5/5] ccbcb

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

Step 3

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