Morphocompletion for #5277 ⟨a, b | aabbbba=bbaa

Solved by morph:4/0,3/0,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabbbba ⇒ bbaa
2. aabbbbbbaa ⇒ bbabbaa
3. aabbbbbbbbaa ⇒ bbabbbbaa
4. aabbbbbbabbaa ⇒ bbabbabbaa
5. aabbbbbbbbbbaa ⇒ bbabbbbbbaa
6. aabbbbbbabbbbaa ⇒ bbabbabbbbaa
7. aabbbbbbbbabbaa ⇒ bbabbbbabbaa
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] bb, [2/1] aa, [2/2] ba
Length 3:[3/0] bbb, [3/1] aab, [3/2] baa
Length 4:[4/0] bbbb, [4/1] aabb, [4/2] bbaa
Length 5:[5/0] bbbbb, [5/1] aabbb, [5/2] bbbba
Length 6:[6/0] bbbbbb, [6/1] aabbbb, [6/2] bbbbaa

Considering [length 4 / frequency 0] bbbb=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aacaca ⇒ caa
2. cb ⇒ bc
3. bbaa ⇒ aaca
4. aacaccaa ⇒ cacaa
5. bbcaa ⇒ caaca
6. bbacaa ⇒ aaccaa
7. aacacccaa ⇒ caccaa
8. aacaccacaa ⇒ cacacaa
9. bbccaa ⇒ ccaaca
10. bbaccaa ⇒ aacccaa
11. bbcacaa ⇒ caaccaa
12. bbacacaa ⇒ aaccacaa
13. bbbb ⇒ c
14. aacaccccaa ⇒ cacccaa
15. bbcccaa ⇒ cccaaca
16. bbacccaa ⇒ aaccccaa
17. bbcaccaa ⇒ caacccaa
18. bbacaccaa ⇒ aaccaccaa
19. bbccacaa ⇒ ccaaccaa
20. bbccccaa ⇒ ccccaaca
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] aa, [2/1] bb, [2/2] ca, [2/3] ac, [2/4] cc
Length 3:[3/0] caa, [3/1] bbc, [3/2] aca, [3/3] bba, [3/4] cca
Length 4:[4/0] ccaa, [4/1] bbac, [4/2] acaa, [4/3] aaca, [4/4] bbcc
Length 5:[5/0] cccaa, [5/1] aacac, [5/2] cacaa, [5/3] accaa, [5/4] bbaca

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

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. caa ⇒ d
2. dcad ⇒ cda
3. aacad ⇒ da
4. aadacad ⇒ daa
5. cdaa ⇒ ddacad
6. cada ⇒ dacad
7. dcaca ⇒ cd
8. aacaca ⇒ d
9. dacaca ⇒ cad
10. aacacd ⇒ cad
11. ddacacd ⇒ cdd
12. aadacacd ⇒ dd
13. ccad ⇒ dcacd
14. cb ⇒ bc
15. bbd ⇒ dca
16. bbaa ⇒ aaca
17. bbad ⇒ aacd
18. bbcd ⇒ cdca
19. bbcad ⇒ dcd
20. bbbb ⇒ c
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] ca, [2/1] aa, [2/2] bb, [2/3] ad, [2/4] ac
Length 3:[3/0] aca, [3/1] cad, [3/2] cac, [3/3] acd, [3/4] aac
Length 4:[4/0] cacd, [4/1] caca, [4/2] aaca, [4/3] acac, [4/4] daca
Length 5:[5/0] acacd, [5/1] acaca, [5/2] dacac, [5/3] aacac, [5/4] aadac

Considering [length 3 / frequency 0] aca=e.

Step 4

Rewriting system is complete. See a, b | aabbbba=bbaa.