Morphocompletion for #3826 ⟨a, b | abbbaabbba=b⟩

Back:⟨a, b | abbbaabbba=b⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abbbab ⇒ babbba
2. abbbaabbba ⇒ b
3. abbbbbbb ⇒ bbbbbbba
4. abbbbabbba ⇒ bbbabbbaab
5. abbbaabbbb ⇒ bbbbaabbba
6. abbbbbabbba ⇒ bbbabbbaabb
7. bbbabbbaaabbba ⇒ abbbb
8. bbbabbbaaaabbba ⇒ aabbbb
9. bbbabbbaaaaabbba ⇒ aaabbbb
10. bbbabbbaababbba ⇒ abbbbb
11. bbbabbbaabaabbba ⇒ ababbbb
12. bbbabbbaaababbba ⇒ aabbbbb
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] bb, [2/1] ba, [2/2] ab, [2/3] aa
Length 3:[3/0] bbb, [3/1] bba, [3/2] abb, [3/3] bab, [3/4] aab, [3/5] baa, [3/6] aaa
Length 4:[4/0] bbba, [4/1] abbb, [4/2] bbbb, [4/3] babb, [4/4] bbab, [4/5] bbaa, [4/6] aabb
Length 5:[5/0] abbba, [5/1] bbbab, [5/2] babbb, [5/3] abbbb, [5/4] bbabb, [5/5] bbbaa, [5/6] aabbb
Length 6:[6/0] bbbabb, [6/1] babbba, [6/2] aabbba, [6/3] abbbaa, [6/4] bbabbb, [6/5] abbbbb, [6/6] bbbaab
Length 7:[7/0] bbbabbb, [7/1] bbabbba, [7/2] abbbaab, [7/3] babbbaa, [7/4] aaabbba, [7/5] ababbba, [7/6] baabbba

Considering [length 4 / frequency 0] bbba=c.

Step 2

Rewriting system is complete. See ⟨a, b | abbbaabbba=b⟩.