Morphocompletion for #5045 ⟨a, b | aaababa=baaa

Solved by morph:3/2,4/1,4/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aaababa ⇒ baaa
2. aaababbaaa ⇒ baabaaa
3. aaababbbaaa ⇒ baabbaaa
4. aaababbabaaa ⇒ baababaaa
5. aaababbbbaaa ⇒ baabbbaaa
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] aa, [2/1] ba, [2/2] ab
Length 3:[3/0] aaa, [3/1] aba, [3/2] bab
Length 4:[4/0] aaab, [4/1] baaa, [4/2] abab
Length 5:[5/0] aaaba, [5/1] bbaaa, [5/2] aabab
Length 6:[6/0] aaabab, [6/1] bbbaaa, [6/2] aababb

Considering [length 3 / frequency 2] bab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. baaa ⇒ aaaca
2. aaacaaca ⇒ caaa
3. bacaaa ⇒ caaaaca
4. baacaaa ⇒ aaaccaaa
5. cab ⇒ bac
6. bab ⇒ c
7. aaacacaaca ⇒ bcaaa
8. aaacaaccaaa ⇒ caacaaa
9. baaccaaa ⇒ aaacccaaa
10. bacacaaa ⇒ cacaaaaca
11. baacacaaa ⇒ aaaccacaaa
12. bacaacaaa ⇒ caaaaccaaa
13. aaacaabac ⇒ caaab
14. baabcaaa ⇒ aaacbcaaa
15. aaacacacaaca ⇒ bbcaaa
16. baacccaaa ⇒ aaaccccaaa
17. baabccaaa ⇒ aaacbccaaa
18. aaacacaabac ⇒ bcaaab
19. baacbcaaa ⇒ aaaccbcaaa
20. baabbcaaa ⇒ aaacbbcaaa
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] aa, [2/1] ca, [2/2] ac, [2/3] ba, [2/4] ab
Length 3:[3/0] aaa, [3/1] aca, [3/2] caa, [3/3] baa, [3/4] aac
Length 4:[4/0] caaa, [4/1] aaca, [4/2] acaa, [4/3] aaac, [4/4] baac
Length 5:[5/0] aaaca, [5/1] acaaa, [5/2] ccaaa, [5/3] caaca, [5/4] bcaaa

Considering [length 4 / frequency 1] aaca=d.

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aaadacd ⇒ dad
2. aaaaacd ⇒ daa
3. daca ⇒ aacd
4. aaca ⇒ d
5. caad ⇒ adacd
6. caaa ⇒ aaacd
7. bad ⇒ adca
8. bdad ⇒ addacd
9. baad ⇒ aaacd
10. bdaa ⇒ adaacd
11. baaa ⇒ ad
12. aaacdca ⇒ cad
13. adaacdca ⇒ bdd
14. aabac ⇒ db
15. aaadabac ⇒ daab
16. bcad ⇒ adcdca
17. aacdb ⇒ dabac
18. cab ⇒ bac
19. bab ⇒ c
20. bbdd ⇒ adccad
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] aa, [2/1] ca, [2/2] ad, [2/3] ac, [2/4] ba
Length 3:[3/0] aaa, [3/1] acd, [3/2] aac, [3/3] aad, [3/4] bac
Length 4:[4/0] aacd, [4/1] abac, [4/2] cdca, [4/3] aaad, [4/4] aaac
Length 5:[5/0] acdca, [5/1] aaacd, [5/2] aaada, [5/3] dabac, [5/4] aacdc

Considering [length 4 / frequency 0] aacd=e.

Step 4

Rewriting system is complete. See a, b | aaababa=baaa.