Morphocompletion for #4192 ⟨a, b | aabbbaaab=ba

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. aabbbaaab ⇒ ba
2. aabbbaba ⇒ babbaaab
3. babbaaababbbaaab ⇒ aabbbabba
4. aabbbaaaaabbbabba ⇒ baabbaaababbbaaab
5. babbaaababbbaba ⇒ aabbbabbabbaaab
6. baabbaaababbbaaababbbaaab ⇒ aabbbaaaaabbbabbba
7. aabbbaaaaabbbaaaaabbbabbba ⇒ baaabbaaababbbaaababbbaaab
8. babbaaababbbaaaaabbbabba ⇒ aabbbabbaabbaaababbbaaab
9. baabbaaababbbaaababbbaba ⇒ aabbbaaaaabbbabbbabbaaab
10. baabbaaababbbaaabaababbbaaab ⇒ aabbbaaaaabbaabbbabba
11. aabbbaaaaabbbaaaaabbaabbbabba ⇒ baaabbaaababbbaaabaababbbaaab
12. baabbaaababbbaaabaababbbaba ⇒ aabbbaaaaabbaabbbabbabbaaab
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ba, [2/1] aa, [2/2] ab, [2/3] bb
Length 3:[3/0] aab, [3/1] bba, [3/2] abb, [3/3] aaa
Length 4:[4/0] aaab, [4/1] bbba, [4/2] abbb, [4/3] bbaa
Length 5:[5/0] abbba, [5/1] bbaaa, [5/2] baaab, [5/3] aabbb
Length 6:[6/0] bbaaab, [6/1] aabbba, [6/2] bbbaaa, [6/3] abbbaa

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. aab ⇒ c
2. cbbac ⇒ ba
3. aacbbabc ⇒ cabbacab
4. aacbbaba ⇒ cabbac
5. babbac ⇒ cbbaba
6. cabbacabbbac ⇒ aacbbabba
7. cbbaaacbbabbc ⇒ bcbacabbbacab
8. cbbaaacbcbbabc ⇒ bcbacabbbaccab
9. cbbaaacbbabba ⇒ bcbacabbbac
10. cbbaaacbcbbaba ⇒ bcbacabbbacc
11. ccbacabbbacabbbac ⇒ aacbbaaacbbabbba
12. cbbabcbacabbbac ⇒ babbaaacbbabba
13. ccbacabbbaccabbbac ⇒ aacbbaaacbcbbabba
14. bcbacabbbacabbbac ⇒ cbbaaacbbabbba
15. bcbacabbbaccabbbac ⇒ cbbaaacbcbbabba
16. cbbababbaaacbbabbc ⇒ babbabcbacabbbacab
17. cbbababbaaacbbabba ⇒ babbabcbacabbbac
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ba, [2/1] bb, [2/2] ac, [2/3] cb, [2/4] ab, [2/5] aa, [2/6] bc
Length 3:[3/0] bba, [3/1] bac, [3/2] cbb, [3/3] abb, [3/4] bab, [3/5] cab, [3/6] bbb
Length 4:[4/0] cbba, [4/1] bbac, [4/2] cabb, [4/3] bbab, [4/4] bbba, [4/5] abbb, [4/6] aacb
Length 5:[5/0] bbbac, [5/1] cbbab, [5/2] abbba, [5/3] cabbb, [5/4] babba, [5/5] cbbaa, [5/6] bacab

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

Step 3

Rewriting system is complete. See a, b | aabbbaaab=ba.