Morphocompletion for #1533 ⟨a, b | abbaabbaab=1⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. baab ⇒ abba
2. baaabba ⇒ abbaaab
3. baaababba ⇒ abbaaabab
4. baaabababba ⇒ abbaaababab
5. baaababababba ⇒ abbaaabababab
6. abbaaaaabababbaba ⇒ abbaaaa
7. abbaaaababababbaba ⇒ abbaaaab
8. abbabaaaabababbaba ⇒ abbabaaa
...

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] bba, [3/3] baa
Length 4:[4/0] abba, [4/1] baba, [4/2] abab, [4/3] baaa
Length 5:[5/0] babba, [5/1] baaab, [5/2] bbaba, [5/3] babab
Length 6:[6/0] ababba, [6/1] abbaba, [6/2] ababab, [6/3] baaaba

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ccaccc ⇒ a
2. cccbcc ⇒ b
3. ccaca ⇒ aaccc
4. accbcc ⇒ 1
5. cab ⇒ abc
6. cacb ⇒ acbc
7. caccb ⇒ accbc
8. cacccbc ⇒ ab
9. ba ⇒ c
10. bcca ⇒ accb
11. bccac ⇒ accbc
12. cbca ⇒ bcac
13. cbcca ⇒ accbc
14. cccbb ⇒ bcbcc
15. aaccb ⇒ ccac
16. abca ⇒ cac
17. abcca ⇒ ccac
18. accbb ⇒ cbcc
19. bccaa ⇒ accc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] cc, [2/1] ca, [2/2] cb, [2/3] bc, [2/4] ac, [2/5] ab, [2/6] bb
Length 3:[3/0] bcc, [3/1] cca, [3/2] cac, [3/3] ccb, [3/4] cbc, [3/5] acc, [3/6] ccc
Length 4:[4/0] bcca, [4/1] accb, [4/2] cbcc, [4/3] ccac, [4/4] cccb, [4/5] cacc, [4/6] ccbb
Length 5:[5/0] cccbc, [5/1] ccbcc, [5/2] caccc, [5/3] accbc, [5/4] ccacc, [5/5] acccb, [5/6] bcbcc

Considering [length 4 / frequency 0] bcca=d.

Step 3

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