Morphocompletion for #1099 ⟨a, b | aabbba=baa

Solved by morph:3/0,4/0,5/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabbba ⇒ baa
2. aabbbbaa ⇒ babaa
3. aabbbbbaa ⇒ babbaa
4. aabbbbabaa ⇒ bababaa
5. aabbbbbbaa ⇒ babbbaa
6. aabbbbabbaa ⇒ bababbaa
7. aabbbbbabaa ⇒ babbabaa
...

Collecting factors up to length 7, frequency 3:

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

Considering [length 3 / frequency 0] bbb=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aacacaca ⇒ caa
2. aacacaccaa ⇒ cacaa
3. aacacacccaa ⇒ caccaa
4. aacacaccacaa ⇒ cacacaa
5. aacacaccccaa ⇒ cacccaa
6. baa ⇒ aaca
7. bc ⇒ cb
8. bacaa ⇒ aaccaa
9. baccaa ⇒ aacccaa
10. bacacaa ⇒ aaccacaa
11. bacccaa ⇒ aaccccaa
12. bacaccaa ⇒ aaccaccaa
13. baccacaa ⇒ aacccacaa
14. bacacacaa ⇒ aaccacacaa
15. baccccaa ⇒ aacccccaa
16. bacacccaa ⇒ aaccacccaa
17. baccaccaa ⇒ aacccaccaa
18. bacccacaa ⇒ aaccccacaa
19. bacccccaa ⇒ aaccccccaa
20. bbb ⇒ c
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] aa, [2/1] ca, [2/2] ac, [2/3] ba, [2/4] cc
Length 3:[3/0] caa, [3/1] bac, [3/2] aca, [3/3] cac, [3/4] cca
Length 4:[4/0] ccaa, [4/1] acac, [4/2] bacc, [4/3] caca, [4/4] acaa
Length 5:[5/0] cccaa, [5/1] cacaa, [5/2] acaca, [5/3] aacac, [5/4] baccc

Considering [length 4 / frequency 0] ccaa=d.

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ccaa ⇒ d
2. aacacad ⇒ cacaa
3. dcacaca ⇒ cd
4. aacacaca ⇒ caa
5. aacacacd ⇒ cad
6. ccacaa ⇒ dacacaca
7. cccad ⇒ dcacacd
8. ccacad ⇒ dacacacd
9. aacacaccd ⇒ cacd
10. bd ⇒ dca
11. baa ⇒ aaca
12. bad ⇒ aacd
13. bc ⇒ cb
14. bacd ⇒ aaccd
15. bacaa ⇒ aad
16. bacad ⇒ aaccad
17. baccd ⇒ aacccd
18. bacacd ⇒ aaccacd
19. bacccd ⇒ aaccccd
20. bbb ⇒ c
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] ca, [2/1] ac, [2/2] ba, [2/3] aa, [2/4] cc
Length 3:[3/0] aca, [3/1] bac, [3/2] cac, [3/3] cad, [3/4] aac
Length 4:[4/0] caca, [4/1] acac, [4/2] aaca, [4/3] baca, [4/4] acad
Length 5:[5/0] acaca, [5/1] aacac, [5/2] ccaca, [5/3] cacac, [5/4] cacad

Considering [length 5 / frequency 0] acaca=e.

Step 4

Rewriting system is complete. See a, b | aabbba=baa.