Morphocompletion for #2528 ⟨a, b | aabbaa=baab

Solved by morph:5/1,2/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. aabbaa ⇒ baab
2. aabbbaab ⇒ baabbbaa
3. aabbabaab ⇒ baababbaa
4. aabbbbaab ⇒ bbaabbbaaaa
5. aabbabbaab ⇒ baababbaabaa
6. aabbbbbaab ⇒ bbaabbbaaaabaa
7. aabbabbbaab ⇒ baababbaabaabaa
8. aabbbbbbaab ⇒ bbaabbbaaaabaabaa
9. aabbabbbbaab ⇒ baababbaabaabaabaa
10. aabbbbbbbaab ⇒ bbaabbbaaaabaabaabaa
11. aabbabbbbbaab ⇒ baababbaabaabaabaabaa
12. aabbbbbbbbaab ⇒ bbaabbbaaaabaabaabaabaa
13. aabbabbbbbbaab ⇒ baababbaabaabaabaabaabaa
14. aabbbbbbbbbaab ⇒ bbaabbbaaaabaabaabaabaabaa
15. aabbabbbbbbbaab ⇒ baababbaabaabaabaabaabaabaa
16. aabbbbbbbbbbaab ⇒ bbaabbbaaaabaabaabaabaabaabaa
17. aabbabbbbbbbbaab ⇒ baababbaabaabaabaabaabaabaabaa
18. aabbabbbbbbbbbaabbbaaaa ⇒ baababbbaabbbbbbbbbaa
19. aabbbbbbbbbbbbaabbbaaaa ⇒ bbbaabbbaaaabbbbbbbbbaa
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] bb, [2/1] aa, [2/2] ab, [2/3] ba
Length 3:[3/0] bbb, [3/1] aab, [3/2] abb, [3/3] bba
Length 4:[4/0] bbbb, [4/1] aabb, [4/2] baab, [4/3] bbaa
Length 5:[5/0] bbbbb, [5/1] bbaab, [5/2] aabbb, [5/3] aabba
Length 6:[6/0] bbbbbb, [6/1] bbbaab, [6/2] aabbab, [6/3] aabbbb

Considering [length 5 / frequency 1] bbaab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aacaa ⇒ c
2. aacc ⇒ ccaa
3. aacac ⇒ cacaa
4. bc ⇒ cbaa
5. baac ⇒ cb
6. aabbaa ⇒ baab
7. aacbaab ⇒ cbbaa
8. aacabaab ⇒ cabbaa
9. bbaaaac ⇒ caabb
10. aabbac ⇒ baabacaa
11. bbaab ⇒ c
12. baabb ⇒ aac
13. aabbabaab ⇒ baababbaa
14. cbbaaaabb ⇒ aacbaaaac
15. cabbaaaabb ⇒ aacabaaaac
16. cbbbaaaabb ⇒ baaaacbaaaac
17. cbabbaaaabb ⇒ baaaacabaaaac
18. baababbaaaabb ⇒ aabbabaaaac
19. cbbbbaaaabb ⇒ baabaaaacbaaaac
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] aa, [2/1] bb, [2/2] ab, [2/3] ba, [2/4] ac, [2/5] cb, [2/6] ca
Length 3:[3/0] aab, [3/1] abb, [3/2] baa, [3/3] aac, [3/4] aaa, [3/5] bba, [3/6] cbb
Length 4:[4/0] aabb, [4/1] baab, [4/2] bbaa, [4/3] baaa, [4/4] aaaa, [4/5] abba, [4/6] aaab
Length 5:[5/0] aaabb, [5/1] bbaaa, [5/2] baaaa, [5/3] aabba, [5/4] aaaab, [5/5] abbaa, [5/6] abaab

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

Step 3

Rewriting system is complete. See a, b | aabbaa=baab.