Morphocompletion for #1777 ⟨a, b | ababaaaab=a

Solved by morph:2/1,3/3,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aaaaabb ⇒ ababaaa
2. aabaaaab ⇒ ababaaaa
3. ababaaaab ⇒ a
4. aaaaababbab ⇒ ababaaababa
5. aaaaabababbab ⇒ aaaaababbbaba
6. ababaaababaaaa ⇒ aaaaab
7. ababaaaaaaaab ⇒ aabaaababaaaa
8. aaaaababbbabaaaa ⇒ aaaaabab
9. ababaaababababaaa ⇒ aaaaababb
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] aa, [2/1] ab, [2/2] ba
Length 3:[3/0] aaa, [3/1] aba, [3/2] bab
Length 4:[4/0] aaaa, [4/1] abab, [4/2] aaab
Length 5:[5/0] ababa, [5/1] aaaaa, [5/2] aaaab
Length 6:[6/0] aaaaab, [6/1] ababaa, [6/2] babaaa

Considering [length 2 / frequency 1] ab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. ab ⇒ c
2. ccaaac ⇒ a
3. acaaac ⇒ ccaaaa
4. aaaacb ⇒ ccaaa
5. aaaaccbb ⇒ ccaaccccaac
6. aaaaccbc ⇒ ccaacca
7. ccaaccccaaa ⇒ aaaaccb
8. ccaaccccaacca ⇒ aaaacccbc
9. aaaacacbb ⇒ ccaaccaccaac
10. aaaacacbc ⇒ ccaaccaa
11. aaaaccacbc ⇒ aaaacccbca
12. ccaaccaaaa ⇒ aaaac
13. ccaaccaccaaa ⇒ aaaacacb
14. aaaacccbcaac ⇒ aaaaccb
15. aaaacccbcaaa ⇒ aaaacc
16. aaaaccccbcaaa ⇒ aaaaccc
17. ccaaaaaaac ⇒ acaaccaaaa
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] aa, [2/1] cc, [2/2] ac, [2/3] ca, [2/4] cb
Length 3:[3/0] aaa, [3/1] aac, [3/2] cca, [3/3] caa, [3/4] acc
Length 4:[4/0] aaaa, [4/1] ccaa, [4/2] aaac, [4/3] caaa, [4/4] aacc
Length 5:[5/0] aaaac, [5/1] ccaaa, [5/2] ccaac, [5/3] aaacc, [5/4] aacca

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

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ab ⇒ c
2. ccdad ⇒ da
3. cac ⇒ db
4. ddac ⇒ dbdbdad
5. cdac ⇒ a
6. dbdac ⇒ d
7. cadb ⇒ dbac
8. dbdadb ⇒ dac
9. cadc ⇒ dbcdadb
10. daa ⇒ dbdad
11. caa ⇒ d
12. dbaa ⇒ cad
13. cada ⇒ dbcdad
14. dbdada ⇒ dcdad
15. adad ⇒ cdcdada
16. acdad ⇒ cdada
17. aac ⇒ cdadb
18. adac ⇒ cdbdad
19. aadb ⇒ cdadbac
20. aaa ⇒ cdad
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] ad, [2/1] da, [2/2] aa, [2/3] ac, [2/4] db
Length 3:[3/0] ada, [3/1] dac, [3/2] dad, [3/3] adb, [3/4] cad
Length 4:[4/0] dbda, [4/1] cdad, [4/2] acda, [4/3] dada, [4/4] dadb
Length 5:[5/0] dbdad, [5/1] bdada, [5/2] bdadb, [5/3] adbac, [5/4] dadba

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

Step 4

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