Morphocompletion for #1249 ⟨a, b | abaab=baba

Solved by morph:5/0,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. baba ⇒ abaab
2. abaabba ⇒ baabaab
3. abaababba ⇒ bbaabaab
4. abaaabaabbba ⇒ bbbaabaab
5. abaabaabaabbba ⇒ bbbbaabaab
6. abaaabaaabaababbba ⇒ bbbbbaabaab
7. abaabaabaaabaababbba ⇒ bbbbbbaabaab
8. abaaabaaabaabaabaababbba ⇒ bbbbbbbaabaab
9. abaabaabaaabaabaabaababbba ⇒ bbbbbbbbaabaab
10. abaaabaaabaabaabaabaabaababbba ⇒ bbbbbbbbbaabaab
11. abaabaabaaabaabaabaabaabaababbba ⇒ bbbbbbbbbbaabaab
12. abaaabaaabaabaabaabaabaabaabaababbba ⇒ bbbbbbbbbbbaabaab
13. abaabaabaaabaabaabaabaabaabaabaababbba ⇒ bbbbbbbbbbbbaabaab
14. abaaabaaabaabaabaabaabaabaabaabaabaababbba ⇒ bbbbbbbbbbbbbaabaab
15. abaabaabaaabaabaabaabaabaabaabaabaabaababbba ⇒ bbbbbbbbbbbbbbaabaab
16. abaaabaaabaabaabaabaabaabaabaabaabaabaabaababbba ⇒ bbbbbbbbbbbbbbbaabaab
17. abaabaabaaabaabaabaabaabaabaabaabaabaabaabaababbba ⇒ bbbbbbbbbbbbbbbbaabaab
18. abaaabaaabaabaabaabaabaabaabaabaabaabaabaabaabaababbba ⇒ bbbbbbbbbbbbbbbbbaabaab
19. abaabaabaaabaabaabaabaabaabaabaabaabaabaabaabaabaababbba ⇒ bbbbbbbbbbbbbbbbbbaabaab
20. abaaabaaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaababbba ⇒ bbbbbbbbbbbbbbbbbbbaabaab
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ba, [2/1] ab, [2/2] aa, [2/3] bb
Length 3:[3/0] aba, [3/1] baa, [3/2] aab, [3/3] bba
Length 4:[4/0] abaa, [4/1] aaba, [4/2] baab, [4/3] bbba
Length 5:[5/0] abaab, [5/1] aabaa, [5/2] baaba, [5/3] abbba
Length 6:[6/0] abaaba, [6/1] aabaab, [6/2] baabaa, [6/3] abaaab

Considering [length 5 / frequency 0] abaab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. acba ⇒ caab
2. accba ⇒ caacab
3. acccba ⇒ caacacab
4. accccba ⇒ caacacacab
5. bc ⇒ cab
6. bac ⇒ cba
7. abaac ⇒ caba
8. abaab ⇒ c
9. caabba ⇒ acc
10. caacabba ⇒ accc
11. caacacabba ⇒ acccc
12. caacacacabba ⇒ accccc
13. baba ⇒ c
14. abaaabaaacc ⇒ ccaaabba
15. cbaaabba ⇒ baacc
16. cabaaabba ⇒ abaaacc
17. cbaaacabba ⇒ baaccc
18. cabaaacabba ⇒ abaaaccc
19. cbaaacacabba ⇒ baacccc
20. cabbaaabba ⇒ bbaacc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ba, [2/1] ca, [2/2] ab, [2/3] ac, [2/4] aa, [2/5] bb, [2/6] cb
Length 3:[3/0] bba, [3/1] cba, [3/2] cab, [3/3] abb, [3/4] aba, [3/5] baa, [3/6] aca
Length 4:[4/0] abba, [4/1] abaa, [4/2] cabb, [4/3] baaa, [4/4] cbaa, [4/5] ccba, [4/6] acab
Length 5:[5/0] cabba, [5/1] aabba, [5/2] acabb, [5/3] cbaaa, [5/4] caaca, [5/5] abaaa, [5/6] baaab

Considering [length 3 / frequency 0] bba=d.

Step 3

Rewriting system is complete. See a, b | abaab=baba.