Morphocompletion for #5874 ⟨a, b | ababab=baaba

Solved by morph:2/0,5/1,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. abbaaba ⇒ baabaab
2. ababab ⇒ baaba
3. baabaabbab ⇒ abbabaaba
4. baabaababbab ⇒ ababbabaaba
5. abbabaabaabab ⇒ baabaabbbaaba
6. ababaabbabaaba ⇒ baabaaabaabbab
7. baabaabbbaabaab ⇒ abbabaababaaba
8. baabaababaabbab ⇒ abbaaabbabaaba
9. ababbaaabbabaaba ⇒ baababaabaaabbab
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] ab, [2/1] ba, [2/2] aa
Length 3:[3/0] aba, [3/1] baa, [3/2] aab
Length 4:[4/0] baab, [4/1] aaba, [4/2] abab
Length 5:[5/0] baaba, [5/1] abaab, [5/2] abbab
Length 6:[6/0] baabaa, [6/1] abaaba, [6/2] aabaab

Considering [length 2 / frequency 0] ab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. caaccc ⇒ acccca
2. caca ⇒ accc
3. accccaaca ⇒ caaccaccc
4. caaccacccca ⇒ accccaaaccc
5. baaccc ⇒ cccca
6. baca ⇒ ccc
7. baaccaccc ⇒ ccccaaca
8. baaccaccaccc ⇒ ccccaacaaca
9. baccaaccaccc ⇒ cccccccaaca
10. cccb ⇒ bacc
11. ab ⇒ c
12. caacbacc ⇒ accccacb
13. caaccbacc ⇒ accccaccb
14. baacbacc ⇒ ccccacb
15. baaccbacc ⇒ ccccaccb
16. baaccacbacc ⇒ ccccaacacb
17. ccccacbcb ⇒ baacbcacc
18. ccccaccbcb ⇒ baaccbcacc
19. caacbacbacc ⇒ accccacbccb
20. baacbacbacc ⇒ ccccacbccb
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] cc, [2/1] ac, [2/2] ca, [2/3] ba, [2/4] cb
Length 3:[3/0] acc, [3/1] ccc, [3/2] baa, [3/3] aac, [3/4] bac
Length 4:[4/0] bacc, [4/1] baac, [4/2] accc, [4/3] caac, [4/4] cbac
Length 5:[5/0] cbacc, [5/1] baacc, [5/2] acbac, [5/3] caacc, [5/4] cccca

Considering [length 5 / frequency 1] baacc=d.

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. cccad ⇒ dcacc
2. cdcad ⇒ ddccc
3. cdcacc ⇒ dcd
4. dcca ⇒ cccaccc
5. cccca ⇒ dc
6. caad ⇒ acccacc
7. caacc ⇒ ad
8. caacdc ⇒ adccca
9. caca ⇒ accc
10. caacad ⇒ adaacc
11. baad ⇒ cccacc
12. baacc ⇒ d
13. baacdc ⇒ dccca
14. baca ⇒ ccc
15. baacad ⇒ daacc
16. baacaccc ⇒ daca
17. dcb ⇒ ccccc
18. cccb ⇒ bacc
19. ab ⇒ c
20. baacbacc ⇒ dccb
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] ca, [2/1] cc, [2/2] ba, [2/3] ac, [2/4] ad
Length 3:[3/0] baa, [3/1] acc, [3/2] ccc, [3/3] caa, [3/4] aac
Length 4:[4/0] baac, [4/1] caac, [4/2] acad, [4/3] aacc, [4/4] cdca
Length 5:[5/0] baaca, [5/1] aacad, [5/2] aacdc, [5/3] baacb, [5/4] cbacc

Considering [length 3 / frequency 0] baa=e.

Step 4

Rewriting system is complete. See a, b | ababab=baaba.