Morphocompletion for #71 ⟨a, b | ababba=1⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. baab ⇒ abba
2. abab ⇒ bbaa
3. babbaa ⇒ 1
4. bbaaab ⇒ abbbaa
5. baaabb ⇒ aabbba
6. baaabba ⇒ aabbbaa
7. bbaaaab ⇒ aabbaba
8. bbabaaab ⇒ abbbabaa
9. baaaabbb ⇒ aaabbbba
10. baaabbbbaa ⇒ aabb
11. baaabbbabba ⇒ aabbb
12. aabbabbabaa ⇒ baaab
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ba, [2/1] aa, [2/2] ab, [2/3] bb
Length 3:[3/0] baa, [3/1] aab, [3/2] bba, [3/3] aaa
Length 4:[4/0] baaa, [4/1] aaab, [4/2] aabb, [4/3] bbaa
Length 5:[5/0] baaab, [5/1] aaabb, [5/2] babba, [5/3] aabba
Length 6:[6/0] baaabb, [6/1] baaaab, [6/2] aaabbb, [6/3] bbabaa

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ccbc ⇒ b
2. bbc ⇒ ccccbb
3. bcbc ⇒ ccbb
4. ba ⇒ c
5. cbca ⇒ 1
6. bbca ⇒ ccb
7. bcac ⇒ 1
8. bccac ⇒ cbcca
9. cbccac ⇒ bca
10. cacc ⇒ a
11. bcab ⇒ cbc
12. bccab ⇒ cbcc
13. bcccab ⇒ cbccc
14. acb ⇒ bca
15. acccb ⇒ cbcca
16. abc ⇒ cab
17. bcaa ⇒ acc
18. aacc ⇒ caca
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] bc, [2/1] ca, [2/2] cc, [2/3] ac, [2/4] cb, [2/5] ab, [2/6] aa
Length 3:[3/0] bca, [3/1] cac, [3/2] cbc, [3/3] bcc, [3/4] cab, [3/5] acc, [3/6] cca
Length 4:[4/0] bcca, [4/1] ccab, [4/2] ccac, [4/3] accc, [4/4] bccc, [4/5] cbcc, [4/6] cccb
Length 5:[5/0] cccab, [5/1] bccac, [5/2] bccca, [5/3] cbcca, [5/4] cccbb, [5/5] cbccc, [5/6] ccccb

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

Step 3

Rewriting system is complete. See a, b | ababba=1⟩.