Morphocompletion for #1126 ⟨a, b | ababba=bab⟩

Back:⟨a, b | ababba=bab⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. ababba ⇒ bab
2. aababbbab ⇒ babbba
3. babbabba ⇒ ababbbab
4. babbbaabba ⇒ aababbbbab
5. ababababbbab ⇒ babbbabba
6. ababbbabbba ⇒ babababbbab
7. aaababababbbbab ⇒ babbbabaabba
8. ababaababbbbab ⇒ babbbbaabba
9. babbbabbaabba ⇒ ababababbbbab
...

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] bab, [3/1] bba, [3/2] aba, [3/3] abb, [3/4] bbb, [3/5] aab, [3/6] baa
Length 4:[4/0] babb, [4/1] abab, [4/2] bbab, [4/3] abba, [4/4] bbba, [4/5] abbb, [4/6] baba
Length 5:[5/0] bbbab, [5/1] babbb, [5/2] ababb, [5/3] babba, [5/4] abbba, [5/5] ababa, [5/6] babab
Length 6:[6/0] babbba, [6/1] abbbab, [6/2] ababbb, [6/3] ababab, [6/4] bbbbab, [6/5] baabba, [6/6] bbabba
Length 7:[7/0] babbbab, [7/1] abbbbab, [7/2] bbaabba, [7/3] ababbba, [7/4] aababbb, [7/5] abababa, [7/6] bbabbba

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

Step 2

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