Morphocompletion for #1115 ⟨a, b | abaaba=bab⟩

Back:⟨a, b | abaaba=bab⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abaaba ⇒ bab
2. bababa ⇒ ababab
3. baababab ⇒ abababba
4. babbaaba ⇒ abaabbab
5. babbabba ⇒ abbabbab
6. aabababba ⇒ babbab
7. baaababab ⇒ abababbaa
8. abababbaba ⇒ bbabbab
9. baaaababab ⇒ abababbaaa
10. abababbbaba ⇒ bbbabbab
11. aabaabbabbab ⇒ babbabbba
12. abaaabaabbab ⇒ babbbaaba
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ba, [2/1] ab, [2/2] aa, [2/3] bb
Length 3:[3/0] aba, [3/1] bab, [3/2] aab, [3/3] bba
Length 4:[4/0] abab, [4/1] baba, [4/2] aaba, [4/3] abba
Length 5:[5/0] babab, [5/1] babba, [5/2] ababa, [5/3] abbab
Length 6:[6/0] ababab, [6/1] babbab, [6/2] aababa, [6/3] bbabba

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ccca ⇒ accc
2. bc ⇒ cca
3. cb ⇒ acc
4. bacc ⇒ ccab
5. accacca ⇒ cccc
6. accab ⇒ ccc
7. aba ⇒ c
8. baaccc ⇒ ccaccaa
9. bacacc ⇒ ccabb
10. bab ⇒ cc
11. baaaccc ⇒ ccaccaaa
12. baacaccc ⇒ ccaccaaca
13. baaccacc ⇒ ccaccaab
14. bacaaccc ⇒ ccaccacaa
15. bacacacc ⇒ ccabbb
16. baaaaccc ⇒ ccaccaaaa
17. baaacaccc ⇒ ccaccaaaca
18. baaaccacc ⇒ ccaccaaab
19. baacaaccc ⇒ ccaccaacaa
20. bacacacacc ⇒ ccabbbb
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] cc, [2/1] ba, [2/2] ac, [2/3] ca, [2/4] aa, [2/5] ab, [2/6] bb
Length 3:[3/0] acc, [3/1] baa, [3/2] ccc, [3/3] cac, [3/4] bac, [3/5] aca, [3/6] aac
Length 4:[4/0] accc, [4/1] cacc, [4/2] baaa, [4/3] baca, [4/4] baac, [4/5] acca, [4/6] acac
Length 5:[5/0] aaccc, [5/1] acacc, [5/2] baaac, [5/3] bacac, [5/4] ccacc, [5/5] baaca, [5/6] baacc

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

Step 3

Rewriting system is complete. See ⟨a, b | abaaba=bab⟩.