Morphocompletion for #1523 ⟨a, b | ababbbabba=1⟩

Solved by morph:6/0,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aababb ⇒ babbaa
2. abbaabab ⇒ babbaaba
3. ababbba ⇒ bbaabab
4. abbbabba ⇒ bbabbaab
5. ababbbab ⇒ bbbabbaa
6. aabbabbaaba ⇒ babbaaaabab
7. bbbabbaaba ⇒ 1
8. abbaabbbaabab ⇒ babbaabaabbba
9. ababbbbabbaa ⇒ bbaababababb
10. ababbbbbaabab ⇒ bbbabbaaabbba
11. bbbabbaabbabbaa ⇒ ababb
12. bbabbaabbbaabab ⇒ abbba
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ab, [2/1] bb, [2/2] ba, [2/3] aa
Length 3:[3/0] bba, [3/1] abb, [3/2] bab, [3/3] aba
Length 4:[4/0] abab, [4/1] abba, [4/2] babb, [4/3] bbaa
Length 5:[5/0] abbaa, [5/1] ababb, [5/2] aabab, [5/3] bbaab
Length 6:[6/0] bbabba, [6/1] ababbb, [6/2] abbaab, [6/3] baabab

Considering [length 6 / frequency 0] bbabba=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. cbcaca ⇒ b
2. cbcacac ⇒ bc
3. cab ⇒ abc
4. babca ⇒ 1
5. bcab ⇒ babc
6. ababc ⇒ 1
7. cbba ⇒ bbac
8. cbcab ⇒ cbabc
9. cbcaabc ⇒ bb
10. bbab ⇒ cbca
11. bbabc ⇒ cbcac
12. bbacb ⇒ ccbca
13. babccab ⇒ bc
14. ababb ⇒ bcaca
15. bbabb ⇒ cbabc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ab, [2/1] bc, [2/2] bb, [2/3] ca, [2/4] cb, [2/5] ba, [2/6] ac
Length 3:[3/0] bab, [3/1] bba, [3/2] cbc, [3/3] abc, [3/4] bca, [3/5] cab, [3/6] abb
Length 4:[4/0] babc, [4/1] cbca, [4/2] babb, [4/3] bbab, [4/4] abab, [4/5] caca, [4/6] bcab
Length 5:[5/0] cbcac, [5/1] bcaca, [5/2] bccab, [5/3] babcc, [5/4] caabc, [5/5] cbcaa, [5/6] cacac

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

Step 3

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