Morphocompletion for #4289 ⟨a, b | ababaabab=ba

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abababa ⇒ baaabab
2. ababaabab ⇒ ba
3. baaababba ⇒ abbaaabab
4. ababaabba ⇒ baabaabab
5. baaabbaaba ⇒ abbaaaabab
6. baabaaababb ⇒ abba
7. baaabbaaabab ⇒ abbaa
8. baaabababba ⇒ abbaabaabab
9. ababaabaabba ⇒ baaabaaababb
10. baaabbaaabba ⇒ abbaaabaabab
11. ababaabaababba ⇒ baaaababbaabab
12. abbaaaababbaba ⇒ baaabbabaaabab
13. abbaaaababbaabab ⇒ baaabbaba
14. baabaababbabaabab ⇒ ababaabbba
...

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] aba, [3/1] baa, [3/2] bab, [3/3] aab
Length 4:[4/0] abab, [4/1] abba, [4/2] aaba, [4/3] baaa
Length 5:[5/0] aabab, [5/1] baaba, [5/2] ababa, [5/3] baaab
Length 6:[6/0] ababba, [6/1] baabab, [6/2] ababaa, [6/3] baaabb

Considering [length 5 / frequency 0] aabab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. accb ⇒ c
2. aba ⇒ cc
3. abc ⇒ ccccb
4. ccbc ⇒ ba
5. abca ⇒ ccccba
6. abcc ⇒ ccba
7. aabba ⇒ cabc
8. ccabba ⇒ ccbacccb
9. accccbba ⇒ cacbc
10. aabbc ⇒ cccbacb
11. accabbc ⇒ cacbacb
12. ccbabc ⇒ abba
13. bacbc ⇒ ccbba
14. abbaa ⇒ bacccba
15. abbac ⇒ bacba
16. bacbacb ⇒ abbc
17. bacccbba ⇒ abbcc
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ba, [2/1] ab, [2/2] cc, [2/3] bc, [2/4] ac, [2/5] cb, [2/6] bb
Length 3:[3/0] bba, [3/1] bac, [3/2] abb, [3/3] ccb, [3/4] acc, [3/5] abc, [3/6] bbc
Length 4:[4/0] abba, [4/1] bacb, [4/2] abbc, [4/3] cbba, [4/4] aabb, [4/5] ccab, [4/6] accc
Length 5:[5/0] ccbba, [5/1] ccabb, [5/2] cbabc, [5/3] cabbc, [5/4] bacba, [5/5] cabba, [5/6] cbacb

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

Step 3

Rewriting system is complete. See a, b | ababaabab=ba.