Morphocompletion for #5401 ⟨a, b | abbabba=abab

Solved by morph:6/1,3/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abbabab ⇒ ababbba
2. abbabba ⇒ abab
3. ababbbaabb ⇒ abababbbba
4. ababbbabab ⇒ ababbabbba
5. ababbbabba ⇒ ababbab
6. ababbabbbaab ⇒ abababbbbaba
7. ababbabbbaba ⇒ ababbbaab
8. abababbbbabba ⇒ abababbbab
9. ababbabbbabba ⇒ abababb
10. abababbabbbbaa ⇒ ababbbaabab
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ab, [2/1] ba, [2/2] bb, [2/3] aa
Length 3:[3/0] bab, [3/1] bba, [3/2] abb, [3/3] aba
Length 4:[4/0] abab, [4/1] babb, [4/2] abba, [4/3] bbab
Length 5:[5/0] ababb, [5/1] babba, [5/2] bbabb, [5/3] babbb
Length 6:[6/0] bbabba, [6/1] ababbb, [6/2] ababba, [6/3] babbba

Considering [length 6 / frequency 1] ababbb=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ccacb ⇒ ccabcc
2. cccabc ⇒ cacb
3. cacba ⇒ cabcca
4. ccaba ⇒ abca
5. ccabca ⇒ abcca
6. cccaba ⇒ cabca
7. cabb ⇒ abbc
8. abccaab ⇒ ccaca
9. cabcaab ⇒ caca
10. abbabc ⇒ caabbb
11. caabbab ⇒ cabca
12. ababbca ⇒ ccaab
13. cababba ⇒ caab
14. ababbb ⇒ c
15. cababbcb ⇒ cccc
16. abbcaab ⇒ ccaba
17. ababbab ⇒ abbca
18. abbabab ⇒ ca
19. abbcabab ⇒ cca
20. abbabba ⇒ abab
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ab, [2/1] ca, [2/2] ba, [2/3] bb, [2/4] cc, [2/5] bc, [2/6] cb
Length 3:[3/0] abb, [3/1] bab, [3/2] aba, [3/3] cab, [3/4] bba, [3/5] cca, [3/6] abc
Length 4:[4/0] abba, [4/1] abab, [4/2] caba, [4/3] caab, [4/4] bbab, [4/5] babb, [4/6] abbc
Length 5:[5/0] abbab, [5/1] ababb, [5/2] abbca, [5/3] cabab, [5/4] babba, [5/5] bcaab, [5/6] cabca

Considering [length 3 / frequency 1] bab=d.

Step 3

Rewriting system is complete. See a, b | abbabba=abab.