Morphocompletion for #5855 ⟨a, b | abaaba=baaab

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: workTotalLimit

#Rule
1. abaaba ⇒ baaab
2. ababaaab ⇒ baaababa
3. abbaabaaab ⇒ baabaaababaaa
4. abbaaabaab ⇒ baabaaababaa
5. abaabbaaab ⇒ baaabbaaba
6. abbbaabaaab ⇒ baabaaababaaaaabaaa
7. abbbaaabaab ⇒ baabaaababaaaaabaa
8. abbabaaabaab ⇒ baabaaababaaaaaba
9. abbbbaabaaab ⇒ baabaaababaaaaabaaaaabaaa
10. abbbbaaabaab ⇒ baabaaababaaaaabaaaaabaa
11. abbbabaaabaab ⇒ baabaaababaaaaabaaaaaba
12. abbbbbaabaaab ⇒ baabaaababaaaaabaaaaabaaaaabaaa
13. abbbbbaaabaab ⇒ baabaaababaaaaabaaaaabaaaaabaa
14. abbbbabaaabaab ⇒ baabaaababaaaaabaaaaabaaaaaba
15. abbbbbabaaabaab ⇒ baabaaababaaaaabaaaaabaaaaabaaaaaba
16. abbbbbbbaabaaabbaabaaaaa ⇒ bbaabaaabbaabaaaabbbbbaa
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ab, [2/1] aa, [2/2] ba, [2/3] bb
Length 3:[3/0] aab, [3/1] baa, [3/2] abb, [3/3] aba
Length 4:[4/0] baab, [4/1] abaa, [4/2] aaab, [4/3] abbb
Length 5:[5/0] baaab, [5/1] abaab, [5/2] abbbb, [5/3] aabaa
Length 6:[6/0] aabaab, [6/1] abaaab, [6/2] abbbbb, [6/3] baaaba

Considering [length 4 / frequency 0] baab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. acaaaaca ⇒ caaac
2. acaaaba ⇒ caaab
3. baac ⇒ caab
4. baaac ⇒ acaaab
5. baaaaca ⇒ acaaaab
6. baab ⇒ c
7. baaab ⇒ aca
8. accaaab ⇒ caaacaabaaaa
9. acacaaab ⇒ caaacaabaaa
10. acaacaaab ⇒ caaacaabaa
11. acaaacaaab ⇒ caaacaaba
12. caaabc ⇒ aacaaacaab
13. caaabac ⇒ acaaacaab
14. caaabb ⇒ aacaaac
15. caaabab ⇒ acaaac
16. bcaaac ⇒ caaacaaaaab
17. bacaaac ⇒ caaacaaaab
18. baaaaacaaac ⇒ caaacb
19. bcaaab ⇒ caaacaaa
20. bacaaab ⇒ caaacaa
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] aa, [2/1] ac, [2/2] ca, [2/3] ab, [2/4] ba, [2/5] bc, [2/6] bb
Length 3:[3/0] aaa, [3/1] aab, [3/2] caa, [3/3] aca, [3/4] aac, [3/5] baa, [3/6] bac
Length 4:[4/0] caaa, [4/1] aaab, [4/2] acaa, [4/3] aaac, [4/4] baaa, [4/5] aaca, [4/6] baca
Length 5:[5/0] caaab, [5/1] acaaa, [5/2] caaac, [5/3] aaaca, [5/4] bacaa, [5/5] bcaaa, [5/6] aaaba

Considering [length 5 / frequency 0] caaab=d.

Step 3

Rewriting system is complete. See a, b | abaaba=baaab.