Morphocompletion for #2503 ⟨a, b | aababa=baaa⟩

Back:⟨a, b | aababa=baaa⟩

Solved by morph:2/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aababa ⇒ baaa
2. aababbaaa ⇒ baabaaa
3. aababbabaaa ⇒ bbaaaaa
4. aababbaabaaa ⇒ baabaabaaa
5. aababbababaaa ⇒ bbaaabaaa
6. aababbbaaaaa ⇒ baabbaaaaa
7. aababbabaabaaa ⇒ bbaaaabaaa
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] aa, [2/1] ba, [2/2] ab, [2/3] bb
Length 3:[3/0] aab, [3/1] aba, [3/2] aaa, [3/3] bab, [3/4] baa, [3/5] bba, [3/6] abb
Length 4:[4/0] aaba, [4/1] baaa, [4/2] abab, [4/3] baba, [4/4] babb, [4/5] abaa, [4/6] abba
Length 5:[5/0] aabab, [5/1] abaaa, [5/2] ababb, [5/3] babba, [5/4] bbaaa, [5/5] babaa, [5/6] ababa
Length 6:[6/0] aababb, [6/1] ababba, [6/2] aabaaa, [6/3] babaaa, [6/4] abbaba, [6/5] babbab, [6/6] baaaaa
Length 7:[7/0] aababba, [7/1] baabaaa, [7/2] babbaba, [7/3] ababbab, [7/4] bbaaaaa, [7/5] ababaaa, [7/6] bbabaaa

Considering [length 2 / frequency 1] ba=c.

Step 2

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