Morphocompletion for #333 ⟨a, b | ababbaba=1⟩

Back:⟨a, b | ababbaba=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. babaa ⇒ aabab
2. bbaba ⇒ ababb
3. baabab ⇒ ababba
4. aababbab ⇒ 1
5. baaaabab ⇒ ababbaaa
6. bbaaabab ⇒ ababbbaa
7. baaaaaabab ⇒ ababbaaaaa
8. baabaaabab ⇒ ababbaabaa
9. bbaaaaabab ⇒ ababbbaaaa
10. aababababbab ⇒ baba
11. baaaaaaaabab ⇒ ababbaaaaaaa
12. baaaabaaabab ⇒ ababbaaaabaa
13. baabaaaaabab ⇒ ababbaabaaaa
14. bbaaaaaaabab ⇒ ababbbaaaaaa
15. bbaaabaaabab ⇒ ababbbaaabaa
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] aa, [2/1] ab, [2/2] ba, [2/3] bb
Length 3:[3/0] bab, [3/1] aaa, [3/2] baa, [3/3] aba
Length 4:[4/0] abab, [4/1] aaba, [4/2] aaaa, [4/3] baaa
Length 5:[5/0] aabab, [5/1] aaaba, [5/2] baaaa, [5/3] aaaaa
Length 6:[6/0] aaabab, [6/1] baaaaa, [6/2] aaaaba, [6/3] aaaaaa

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ca ⇒ ac
2. aa ⇒ c
3. cbab ⇒ babc
4. cbabc ⇒ babcc
5. cbaba ⇒ babac
6. cbabb ⇒ babcb
7. abcbab ⇒ abbabc
8. babcba ⇒ abcbab
9. bbaba ⇒ ababb
10. bbabcbc ⇒ a
11. cbbabcb ⇒ a
12. ababba ⇒ bbabc
13. abbabcb ⇒ 1
14. babbabc ⇒ 1
15. babbabcc ⇒ c
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ba, [2/1] ab, [2/2] cb, [2/3] bc, [2/4] bb, [2/5] cc, [2/6] ac
Length 3:[3/0] bab, [3/1] cba, [3/2] abc, [3/3] bba, [3/4] abb, [3/5] bcb, [3/6] aba
Length 4:[4/0] babc, [4/1] bbab, [4/2] cbab, [4/3] abcb, [4/4] babb, [4/5] abba, [4/6] baba
Length 5:[5/0] bbabc, [5/1] babcb, [5/2] babba, [5/3] abbab, [5/4] abcba, [5/5] ababb, [5/6] cbbab

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

Step 3

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