Morphocompletion for #5664 ⟨a, b | aabaab=baabb

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aabaab ⇒ baabb
2. baabbaab ⇒ aabbaabb
3. aaaabbaabb ⇒ baabbbaab
4. aabbaabbbaab ⇒ bbaabbbaabb
5. aabbaabbbbaab ⇒ bbbaabbbaabb
6. aabbaabbbbbaab ⇒ bbbbaabbbaabb
7. baabbbaabbbaab ⇒ aabbbaabbbaabb
8. aabbaabbbbbbaab ⇒ bbbbbaabbbaabb
9. baabbbaabbbbaab ⇒ aabbbbaabbbaabb
10. aaaabbbaabbbaabb ⇒ baabbbbaabbbaab
11. aabbaabbbbbbbaab ⇒ bbbbbbaabbbaabb
12. baabbbaabbbbbaab ⇒ aabbbbbaabbbaabb
13. aaaabbbbaabbbaabb ⇒ baabbbbaabbbbaab
14. aabbaabbbbbbbbaab ⇒ bbbbbbbaabbbaabb
15. baabbbaabbbbbbaab ⇒ aabbbbbbaabbbaabb
16. aaaabbbbbaabbbaabb ⇒ baabbbbaabbbbbaab
17. aabbaabbbbbbbbbaab ⇒ bbbbbbbbaabbbaabb
18. baabbbaabbbbbbbaab ⇒ aabbbbbbbaabbbaabb
19. aaaabbbbbbaabbbaabb ⇒ baabbbbaabbbbbbaab
20. aabbaabbbbbbbbbbaab ⇒ bbbbbbbbbaabbbaabb
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] bb, [2/1] aa, [2/2] ab, [2/3] ba
Length 3:[3/0] aab, [3/1] bbb, [3/2] baa, [3/3] abb, [3/4] bba, [3/5] aaa, [3/6] aba
Length 4:[4/0] baab, [4/1] aabb, [4/2] bbbb, [4/3] bbaa, [4/4] bbba, [4/5] abbb, [4/6] abba
Length 5:[5/0] bbaab, [5/1] baabb, [5/2] bbbbb, [5/3] bbbaa, [5/4] aabbb, [5/5] aabba, [5/6] bbbba
Length 6:[6/0] bbbaab, [6/1] bbaabb, [6/2] baabbb, [6/3] bbbbbb, [6/4] aabbaa, [6/5] bbbbaa, [6/6] aabbbb
Length 7:[7/0] bbbbaab, [7/1] aabbaab, [7/2] bbbaabb, [7/3] bbaabbb, [7/4] baabbba, [7/5] abbbaab, [7/6] aabbbaa

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

Step 2

Rewriting system is complete. See a, b | aabaab=baabb.