Morphocompletion for #1513 ⟨a, b | ababababba=1⟩

Solved by morph:2/0,5/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. baab ⇒ abba
2. baaabb ⇒ aabbba
3. bbaaab ⇒ abbbaa
4. baaabba ⇒ abbaaab
5. baaababb ⇒ aababbba
6. bababbaa ⇒ abababab
7. baaababba ⇒ abbaaabab
8. babababba ⇒ ababababb
9. aababababb ⇒ 1
10. baaabababab ⇒ abababbabaa
11. aabababababbbaa ⇒ baaab
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] ba, [2/1] ab, [2/2] aa, [2/3] bb
Length 3:[3/0] bab, [3/1] baa, [3/2] aba, [3/3] aab
Length 4:[4/0] abab, [4/1] baba, [4/2] baaa, [4/3] babb
Length 5:[5/0] babab, [5/1] baaab, [5/2] ababb, [5/3] ababa
Length 6:[6/0] ababab, [6/1] bababb, [6/2] bababa, [6/3] baaaba

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. cacccc ⇒ a
2. ccccbc ⇒ b
3. caccca ⇒ aacccc
4. acccbc ⇒ 1
5. cab ⇒ abc
6. cacb ⇒ acbc
7. caccb ⇒ accbc
8. cacccb ⇒ 1
9. ba ⇒ c
10. cbcca ⇒ bccac
11. cbccca ⇒ bcccac
12. ccccbb ⇒ bcccbc
13. aacccb ⇒ caccc
14. abca ⇒ cac
15. abcca ⇒ ccac
16. abccca ⇒ cccac
17. acccbb ⇒ cccbc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] cc, [2/1] ca, [2/2] cb, [2/3] ac, [2/4] bc, [2/5] ab, [2/6] bb
Length 3:[3/0] ccc, [3/1] ccb, [3/2] cca, [3/3] cac, [3/4] acc, [3/5] cbc, [3/6] abc
Length 4:[4/0] cccb, [4/1] cacc, [4/2] accc, [4/3] ccca, [4/4] cccc, [4/5] ccbb, [4/6] bcca
Length 5:[5/0] acccb, [5/1] caccc, [5/2] cccbb, [5/3] bccca, [5/4] ccccb, [5/5] cccbc, [5/6] abccc

Considering [length 5 / frequency 0] acccb=d.

Step 3

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