Morphocompletion for #2249 ⟨a, b | aababba=baa

Solved by morph:6/0,2/0,4/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aababba ⇒ baa
2. aababbbaa ⇒ babaa
3. aababbbbaa ⇒ babbaa
4. aababbbabaa ⇒ bababaa
5. aababbbbbaa ⇒ babbbaa
6. aababbbabbaa ⇒ bababbaa
7. aababbbbabaa ⇒ babbabaa
...

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] baa, [3/2] bbb
Length 4:[4/0] aaba, [4/1] babb, [4/2] bbaa
Length 5:[5/0] aabab, [5/1] ababb, [5/2] babbb
Length 6:[6/0] aababb, [6/1] ababbb, [6/2] bbbbaa

Considering [length 6 / frequency 0] aababb=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. baa ⇒ ca
2. aacbca ⇒ caa
3. aaccacbca ⇒ caaa
4. bac ⇒ cc
5. aacbcc ⇒ cac
6. aaccacbcc ⇒ caac
7. bcaa ⇒ cacbca
8. bcac ⇒ cacbcc
9. aacbbc ⇒ cc
10. cacbbc ⇒ bcc
11. babc ⇒ cbc
12. aacbcbc ⇒ cabc
13. bcabc ⇒ cacbcbc
14. aababb ⇒ c
15. cababb ⇒ bc
16. aacbbbc ⇒ cbc
17. cacbbbc ⇒ bcbc
18. babbc ⇒ cbbc
19. aacbbbbc ⇒ cbbc
20. babbbc ⇒ cbbbc
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] bc, [2/1] aa, [2/2] bb, [2/3] ac, [2/4] ca
Length 3:[3/0] aac, [3/1] bbc, [3/2] bca, [3/3] acb, [3/4] bab
Length 4:[4/0] aacb, [4/1] bbbc, [4/2] babb, [4/3] cacb, [4/4] acbb
Length 5:[5/0] aacbb, [5/1] aacbc, [5/2] cbbbc, [5/3] ababb, [5/4] acbbc

Considering [length 2 / frequency 0] bc=d.

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. aacda ⇒ caa
2. cacda ⇒ daa
3. aacdc ⇒ cac
4. cacdc ⇒ dac
5. aacdd ⇒ cad
6. cacdd ⇒ dad
7. bc ⇒ d
8. aacbd ⇒ cc
9. cacbd ⇒ dc
10. baa ⇒ ca
11. bac ⇒ cc
12. bdc ⇒ dacbd
13. bad ⇒ cd
14. aacbbd ⇒ cd
15. babd ⇒ cbd
16. aababd ⇒ cc
17. cababd ⇒ dc
18. aababb ⇒ c
19. cababb ⇒ d
20. dababb ⇒ bd
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] aa, [2/1] bd, [2/2] ba, [2/3] ca, [2/4] ab
Length 3:[3/0] aac, [3/1] cac, [3/2] bab, [3/3] abb, [3/4] abd
Length 4:[4/0] babb, [4/1] cacd, [4/2] aacd, [4/3] abab, [4/4] babd
Length 5:[5/0] ababb, [5/1] ababd, [5/2] cabab, [5/3] aabab, [5/4] acbbd

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

Step 4

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