Morphocompletion for #533 ⟨a, b | aabba=baa

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aabba ⇒ baa
2. aabbbaa ⇒ babaa
3. aabbbbaa ⇒ babbaa
4. aabbbbbaa ⇒ babbbaa
5. aabbbabaa ⇒ bababaa
6. aabbbbbbaa ⇒ babbbbaa
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] aa, [2/1] bb, [2/2] ba
Length 3:[3/0] aab, [3/1] bbb, [3/2] baa
Length 4:[4/0] aabb, [4/1] bbaa, [4/2] bbbb
Length 5:[5/0] aabbb, [5/1] bbbaa, [5/2] bbbba
Length 6:[6/0] bbbbaa, [6/1] aabbbb, [6/2] aabbba

Considering [length 4 / frequency 0] aabb=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. bac ⇒ cc
2. bcac ⇒ cabcc
3. cabb ⇒ bc
4. babc ⇒ cbc
5. bcabc ⇒ cabcbc
6. aabcc ⇒ cac
7. baa ⇒ ca
8. bcaa ⇒ cabca
9. cabcbbc ⇒ bbcc
10. babbc ⇒ cbbc
11. aabb ⇒ c
12. aabcbc ⇒ cabc
13. aacabcc ⇒ caac
14. aabca ⇒ caa
15. cabcbbbc ⇒ bbcbc
16. babbbc ⇒ cbbbc
17. aabcbbc ⇒ bcc
18. aacabcbc ⇒ caabc
19. aacabca ⇒ caaa
20. aabcbbbc ⇒ bcbc
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] bc, [2/1] aa, [2/2] ab, [2/3] ca, [2/4] bb
Length 3:[3/0] abc, [3/1] aab, [3/2] bbc, [3/3] bca, [3/4] cab
Length 4:[4/0] aabc, [4/1] cabc, [4/2] aaca, [4/3] abcb, [4/4] bbbc
Length 5:[5/0] aacab, [5/1] aabcb, [5/2] cabcb, [5/3] abcbb, [5/4] abcbc

Considering [length 3 / frequency 0] abc=d.

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. cbc ⇒ bd
2. bcd ⇒ cdbc
3. add ⇒ cad
4. adc ⇒ cac
5. acdd ⇒ dad
6. acdc ⇒ dac
7. abc ⇒ d
8. adbc ⇒ cd
9. acdbc ⇒ dd
10. bad ⇒ cd
11. bcad ⇒ cdd
12. bac ⇒ cc
13. bcac ⇒ cdc
14. ada ⇒ caa
15. cabb ⇒ bc
16. abbd ⇒ dbc
17. babd ⇒ cbd
18. baa ⇒ ca
19. bcaa ⇒ cda
20. aabb ⇒ c
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] bc, [2/1] ad, [2/2] ac, [2/3] ba, [2/4] ab
Length 3:[3/0] abb, [3/1] bca, [3/2] acd, [3/3] dbc, [3/4] aab
Length 4:[4/0] acdb, [4/1] cdbc

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

Step 4

Rewriting system is complete. See a, b | aabba=baa.