Morphocompletion for #3265 ⟨a, b | ababbbbabba=1⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. bbbbabbaab ⇒ babbbbabba
2. ababbbbab ⇒ bbbbabbaa
3. bbaababb ⇒ ababbbba
4. aababbb ⇒ bbabbaa
5. babbbbabbaa ⇒ 1
6. bbbabbaaba ⇒ abbbbabbaa
7. abbaabab ⇒ babbaaba
8. babbbbabbabbabbaa ⇒ ababbb
9. ababbbbaaabab ⇒ bbaabbabbaaba
10. bbabbaaabbbbabbaa ⇒ aababb
11. ababbbbaabbbbabbaa ⇒ bbaabab
12. babbbbabbababbaaba ⇒ bbaabab
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] bb, [2/1] ab, [2/2] ba, [2/3] aa
Length 3:[3/0] bab, [3/1] bba, [3/2] abb, [3/3] bbb
Length 4:[4/0] babb, [4/1] bbaa, [4/2] bbab, [4/3] abab
Length 5:[5/0] abbaa, [5/1] babba, [5/2] babbb, [5/3] bbabb
Length 6:[6/0] babbaa, [6/1] bbabba, [6/2] babbbb, [6/3] bbbbab

Considering [length 5 / frequency 0] abbaa=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. bbcbabb ⇒ 1
2. abbbbc ⇒ cbabbb
3. babbbbc ⇒ 1
4. bbcbc ⇒ aa
5. abbaa ⇒ c
6. cbabbbba ⇒ a
7. cbabbbbab ⇒ ab
8. abbcbabb ⇒ a
9. abbcbabbb ⇒ ab
10. cbbcbabb ⇒ c
11. abbac ⇒ cbbaa
12. bbbbcbac ⇒ c
13. cbabbbbabba ⇒ abba
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] bb, [2/1] ab, [2/2] ba, [2/3] cb, [2/4] bc, [2/5] ac, [2/6] aa
Length 3:[3/0] abb, [3/1] bbb, [3/2] bbc, [3/3] bab, [3/4] cba, [3/5] bba, [3/6] bcb
Length 4:[4/0] babb, [4/1] cbab, [4/2] bbcb, [4/3] abbb, [4/4] bbbb, [4/5] abba, [4/6] bcba
Length 5:[5/0] cbabb, [5/1] babbb, [5/2] bbcba, [5/3] bbbbc, [5/4] abbbb, [5/5] abbcb, [5/6] bcbab

Considering [length 5 / frequency 0] cbabb=d.

Step 3

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