Morphocompletion for #1763 ⟨a, b | abaabaaab=a

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aaaabb ⇒ abaaba
2. aaabaaab ⇒ abaabaaa
3. abaabaaab ⇒ a
4. abaabaaaab ⇒ aaaababbaa
5. aaaababaaab ⇒ aaaabaabbaa
6. aaaababbaab ⇒ abaababaaba
7. abaabaaaaaab ⇒ aaababaabaaa
8. abaababaabaaa ⇒ aaaab
9. aaaababbbaabaaa ⇒ aaaabab
10. abaababaababaaba ⇒ aaaababb
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] aa, [2/1] ab, [2/2] ba
Length 3:[3/0] aba, [3/1] aaa, [3/2] aab
Length 4:[4/0] abaa, [4/1] aaab, [4/2] aaba
Length 5:[5/0] abaab, [5/1] aaaab, [5/2] baaba
Length 6:[6/0] abaaba, [6/1] aabaaa, [6/2] 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. ccaac ⇒ a
2. acaac ⇒ ccaaa
3. caab ⇒ abac
4. ccaccaaa ⇒ aaac
5. aaab ⇒ caac
6. aaacb ⇒ ccaca
7. aaaccb ⇒ ccaccccaca
8. aaacccb ⇒ ccaccccaccccaca
9. ccaaaaac ⇒ acaccaaa
10. aaacab ⇒ ccaccccaaa
11. aaacacb ⇒ ccaccaccaca
12. abaab ⇒ c
13. ccaaacaccaaa ⇒ acaaaaac
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] aa, [2/1] ac, [2/2] cc, [2/3] ca, [2/4] ab
Length 3:[3/0] aaa, [3/1] aac, [3/2] cca, [3/3] caa, [3/4] aab
Length 4:[4/0] aaac, [4/1] ccaa, [4/2] caaa, [4/3] caac, [4/4] aaca
Length 5:[5/0] ccaaa, [5/1] aaaca, [5/2] aaacc, [5/3] aacab, [5/4] aaaac

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

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ccaccd ⇒ dc
2. da ⇒ ad
3. ccaac ⇒ a
4. db ⇒ caac
5. dcb ⇒ ccaca
6. ccaccad ⇒ dca
7. ccaadc ⇒ acaccd
8. dcaac ⇒ aaccd
9. aaa ⇒ d
10. ccaccccaca ⇒ dccb
11. acaac ⇒ ccd
12. dcab ⇒ ccaccccd
13. caab ⇒ abac
14. ccaccaad ⇒ dcaa
15. abaab ⇒ c
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] cc, [2/1] ca, [2/2] ac, [2/3] aa, [2/4] ab
Length 3:[3/0] cca, [3/1] caa, [3/2] aac, [3/3] cac, [3/4] aab
Length 4:[4/0] ccac, [4/1] caac, [4/2] ccaa, [4/3] cacc, [4/4] caad
Length 5:[5/0] ccacc, [5/1] ccaad, [5/2] caadc, [5/3] accad, [5/4] cacca

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

Step 4

Rewriting system is complete. See a, b | abaabaaab=a.