Morphocompletion for #3767 ⟨a, b | abababaaab=a

Solved by morph:3/0,2/1,4/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabaaab ⇒ aaaabba
2. aaaabbab ⇒ abababaa
3. aababaaab ⇒ abababaaa
4. abababaaab ⇒ a
5. aaaabbaaab ⇒ aaaabaabba
6. aaaabbaaaab ⇒ aabaaaaabba
7. aaaabbbababaaa ⇒ aaaab
8. abababaabababaa ⇒ aaaabb
...

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] aab, [3/2] aba
Length 4:[4/0] aaab, [4/1] abab, [4/2] baba
Length 5:[5/0] aaaab, [5/1] ababa, [5/2] baaab
Length 6:[6/0] aaaabb, [6/1] abaaab, [6/2] ababaa

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ca ⇒ ac
2. aaa ⇒ c
3. acbcb ⇒ ccbba
4. aabcb ⇒ acbba
5. cbcbaa ⇒ aacbbc
6. aacbba ⇒ cbcb
7. cbabcb ⇒ cbcbba
8. acbcbba ⇒ cbababc
9. aabcbba ⇒ abababc
10. ccbabcb ⇒ ccbcbba
11. acbabcb ⇒ cbababc
12. aababcb ⇒ abababc
13. cbcbbab ⇒ aa
14. cbabcbb ⇒ aa
15. aabababc ⇒ cbabcb
16. cbababaa ⇒ acbcbb
17. abababaa ⇒ aabcbb
18. cbababcb ⇒ c
19. abababcb ⇒ a
20. ccbcbbab ⇒ aac
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] cb, [2/1] ab, [2/2] ba, [2/3] aa, [2/4] bc
Length 3:[3/0] bcb, [3/1] bab, [3/2] aba, [3/3] cba, [3/4] abc
Length 4:[4/0] abcb, [4/1] abab, [4/2] cbab, [4/3] babc, [4/4] cbcb
Length 5:[5/0] babcb, [5/1] ababa, [5/2] bcbba, [5/3] cbabc, [5/4] babab

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

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ca ⇒ ac
2. dddaad ⇒ a
3. cb ⇒ aad
4. cdbd ⇒ dddaa
5. ab ⇒ d
6. dddadddadddc ⇒ cd
7. dddcdaad ⇒ addadddc
8. cddaad ⇒ aadddc
9. aaa ⇒ c
10. dddadddaa ⇒ cdb
11. adaad ⇒ dddadddc
12. addaad ⇒ dddc
13. cdba ⇒ dddadddc
14. cdbb ⇒ dddadddad
15. dddadddadddac ⇒ cda
16. aadddadddc ⇒ cdaad
17. dddacdbc ⇒ cdaa
18. acdaad ⇒ cdddadddc
19. cdbcdb ⇒ adadddaa
20. cdaadb ⇒ aadddaa
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] dd, [2/1] ad, [2/2] da, [2/3] cd, [2/4] aa
Length 3:[3/0] ddd, [3/1] aad, [3/2] dda, [3/3] cdb, [3/4] daa
Length 4:[4/0] ddda, [4/1] daad, [4/2] addd, [4/3] dadd, [4/4] ddad
Length 5:[5/0] dddad, [5/1] cdaad, [5/2] ddaad, [5/3] daddd, [5/4] ddadd

Considering [length 4 / frequency 0] ddda=e.

Step 4

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