Morphocompletion for #4015 ⟨a, b | aaabbaaba=ab⟩

Back:⟨a, b | aaabbaaba=ab⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aaabbaaba ⇒ ab
2. abaabbaaba ⇒ abb
3. aaabbaabb ⇒ ababbaaba
4. abbaabbaaba ⇒ abbb
5. ababbaabaaaba ⇒ aaabbb
6. abaabbaabb ⇒ abbabbaaba
7. aaabbaaaabbb ⇒ abbbaabaaaba
8. abbbaabbaaba ⇒ abbbb
9. abbabbaabaaaba ⇒ abaabbb
10. abbaabbaabb ⇒ abbbabbaaba
11. abbbbaabbaaba ⇒ abbbbb
12. abbbabbaabaaaba ⇒ abbaabbb
13. abbbbbaabbaaba ⇒ abbbbbb
14. abbbbbbaabbaaba ⇒ abbbbbbb
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] ab, [2/1] ba, [2/2] bb, [2/3] aa
Length 3:[3/0] abb, [3/1] aba, [3/2] aab, [3/3] baa, [3/4] bba, [3/5] bbb, [3/6] aaa
Length 4:[4/0] aaba, [4/1] baab, [4/2] abba, [4/3] bbaa, [4/4] aabb, [4/5] abbb, [4/6] aaab
Length 5:[5/0] bbaab, [5/1] abbaa, [5/2] baaba, [5/3] baabb, [5/4] aabba, [5/5] aaabb, [5/6] abbbb
Length 6:[6/0] abbaab, [6/1] bbaaba, [6/2] bbaabb, [6/3] aabbaa, [6/4] baabba, [6/5] aaabba, [6/6] baaaba
Length 7:[7/0] abbaaba, [7/1] abbaabb, [7/2] aabbaab, [7/3] baabbaa, [7/4] bbaabba, [7/5] aaabbaa, [7/6] abaaaba

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

Step 2

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