Morphocompletion for #1262 ⟨a, b | ababa=baab

Solved by morph:3/2,4/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ababa ⇒ baab
2. abbaab ⇒ baabba
3. abbabaab ⇒ babaabbaa
4. abbbabaab ⇒ babaabbaaabaa
5. abbbaabab ⇒ babaabbaaaba
6. abbbbabaab ⇒ babaabbaaabaaabaa
7. abbbbaabab ⇒ babaabbaaabaaaba
8. abbbbbabaab ⇒ babaabbaaabaaabaaabaa
9. abbbbbaabab ⇒ babaabbaaabaaabaaaba
10. abbbbbbabaab ⇒ babaabbaaabaaabaaabaaabaa
11. abbbbbbaabab ⇒ babaabbaaabaaabaaabaaaba
12. abbbbbbbabaab ⇒ babaabbaaabaaabaaabaaabaaabaa
13. abbbbbbbaabab ⇒ babaabbaaabaaabaaabaaabaaaba
14. abbbbbbbbaabab ⇒ babaabbaaabaaabaaabaaabaaabaaaba
15. abbbbbbabaabbabaa ⇒ bbabaabbabaaabbbb
16. abbbbbbbabaabbabaa ⇒ bbabaabbabaaabbbbb
17. abbbbbbbbabaabbabaa ⇒ bbabaabbabaaabbbbbb
18. abbbbbbbbbabaabbabaa ⇒ bbabaabbabaaabbbbbbb
19. abbbbbbbbbbabaabbabaa ⇒ bbabaabbabaaabbbbbbbb
20. abbbbbbbbbbbabaabbabaa ⇒ bbabaabbabaaabbbbbbbbb
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] bb, [2/1] ab, [2/2] ba, [2/3] aa
Length 3:[3/0] bbb, [3/1] abb, [3/2] bab, [3/3] baa
Length 4:[4/0] bbbb, [4/1] abbb, [4/2] baab, [4/3] abaa
Length 5:[5/0] bbbbb, [5/1] abbbb, [5/2] babaa, [5/3] bbaba
Length 6:[6/0] bbbbbb, [6/1] abbbbb, [6/2] bbabaa, [6/3] babaab

Considering [length 3 / frequency 2] bab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. acaaaca ⇒ caac
2. acaaba ⇒ caab
3. bac ⇒ cab
4. baac ⇒ acaab
5. baaaca ⇒ acaaab
6. bab ⇒ c
7. baab ⇒ aca
8. acaacaac ⇒ caacaaca
9. accaab ⇒ caacabaaa
10. acacaab ⇒ caacabaa
11. acaacaab ⇒ caacaba
12. caabc ⇒ acaacab
13. caabb ⇒ acaac
14. bcaac ⇒ caacaaab
15. bcaaaca ⇒ caacaaaab
16. baaaacaac ⇒ acaaacb
17. bcaab ⇒ caacaa
18. caacabb ⇒ acaaaacaac
19. caacabaaab ⇒ acacaac
20. caacabaaaab ⇒ accaac
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] ac, [2/1] ca, [2/2] aa, [2/3] ab, [2/4] ba, [2/5] bc, [2/6] bb
Length 3:[3/0] aca, [3/1] caa, [3/2] aab, [3/3] aac, [3/4] baa, [3/5] aaa, [3/6] bca
Length 4:[4/0] caac, [4/1] caab, [4/2] aaca, [4/3] acaa, [4/4] bcaa, [4/5] baaa, [4/6] aaab
Length 5:[5/0] acaac, [5/1] caaca, [5/2] aaaca, [5/3] acaab, [5/4] aacab, [5/5] baaaa, [5/6] aacaa

Considering [length 4 / frequency 1] caab=d.

Step 3

Rewriting system is complete. See a, b | ababa=baab.