Morphocompletion for #1112 ⟨a, b | abaaba=aab⟩

Back:⟨a, b | abaaba=aab⟩

Solved by morph:3/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abaaba ⇒ aab
2. aababa ⇒ abaaab
3. aaabbaa ⇒ abaaaab
4. ababaaab ⇒ aabba
5. abaaabaab ⇒ aaabba
6. abaaaabba ⇒ aaabab
7. aabbaaba ⇒ aabab
8. aabbaaab ⇒ abaaabba
9. aabbabaaab ⇒ aababba
10. aabbaaaabba ⇒ aaabb
11. aababbaaba ⇒ abaaabb
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] aa, [2/1] ab, [2/2] ba, [2/3] bb
Length 3:[3/0] aab, [3/1] aba, [3/2] baa, [3/3] bba, [3/4] aaa, [3/5] abb, [3/6] bab
Length 4:[4/0] aabb, [4/1] aaab, [4/2] aaba, [4/3] abba, [4/4] abaa, [4/5] bbaa, [4/6] baaa
Length 5:[5/0] aabba, [5/1] baaab, [5/2] abbaa, [5/3] baaba, [5/4] abaaa, [5/5] aaabb, [5/6] aabab
Length 6:[6/0] aabbaa, [6/1] aaabba, [6/2] abaaab, [6/3] bbaaba, [6/4] aababb, [6/5] aabbab, [6/6] abbaab
Length 7:[7/0] abbaaba, [7/1] babaaab, [7/2] aaaabba, [7/3] aabbaaa, [7/4] aababba, [7/5] aabbaab, [7/6] aabbaba

Considering [length 3 / frequency 1] aba=c.

Step 2

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