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

Back:⟨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:

[2/0]bb100lf:100,rf:58,rp:6,rs:6,a:1
[2/1]ab97lf:64,lp:20,ls:13,rf:86,rs:1,lprs:1
[2/2]ba52lf:51,ls:1,rf:99,rp:14,rs:7,lsrp:1
[2/3]aa31lf:25,ls:6,rf:110,rs:6
[3/0]bbb75lf:75,rf:27,rs:6,a:5
[3/1]abb44lf:25,lp:19,rf:25,a:1
[3/2]bab31lf:25,ls:6,rf:24,rp:12,lsrp:6
[3/3]baa31lf:25,ls:6,rf:68,rp:2,rs:6
[4/0]bbbb58lf:58,rf:21,rs:6,a:11
[4/1]abbb34lf:17,lp:17,rf:6,a:5
[4/2]baab26lf:19,ls:7,rf:20,rp:2,rs:1,re:1,lsrp:1
[4/3]abaa24lf:18,ls:6,rf:54,rs:5,a:2
[5/0]bbbbb43lf:43,rf:15,rs:5,a:17
[5/1]abbbb30lf:15,lp:15,rf:6,rs:1,lprs:1,a:11
[5/2]babaa24lf:18,ls:6,rf:24,rp:12,a:2
[5/3]bbaba18lf:18,rf:12,rp:6,a:2
[6/0]bbbbbb30lf:30,rf:10,rs:4,a:23
[6/1]abbbbb26lf:13,lp:13,rf:5,rs:1,lprs:1,a:17
[6/2]bbabaa24lf:18,ls:6,rf:12,rp:6,lsrp:6,a:2
[6/3]babaab18lf:12,ls:6,rf:18,rp:12,lsrp:6,a:2

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:

[2/0]ac32lf:21,lp:6,ls:5,rf:26,rp:9,rs:5,lsrp:2,lprs:1,a:2
[2/1]ca32lf:24,lp:5,ls:3,rf:30,rp:10,rs:2,lsrp:2,a:2
[2/2]aa31lf:31,rf:32,rs:3,a:2
[2/3]ab22lf:14,ls:8,rf:10,rs:7,a:1
[2/4]ba16lf:9,lp:6,ls:1,rf:3,rs:1,a:1
[2/5]bc8lf:4,lp:3,ls:1,a:8
[2/6]bb4lf:2,ls:2,a:9
[3/0]aca23lf:15,lp:5,ls:3,rf:19,rp:8,rs:2,re:1,lsrp:1,lprs:1,a:2
[3/1]caa22lf:17,lp:5,rf:22,rp:9,rs:1,a:4
[3/2]aab17lf:10,ls:7,rf:5,rs:5,a:2
[3/3]aac17lf:13,ls:4,rf:16,rs:5,a:5
[3/4]baa10lf:6,lp:4,rf:2,rs:1,a:2
[3/5]aaa8lf:8,rf:8,rs:1,a:6
[3/6]bca6lf:3,lp:3,a:8
[4/0]caac14lf:8,lp:3,ls:3,rf:14,rp:8,rs:5,re:1,lsrp:2,lprs:3,a:7
[4/1]caab13lf:7,lp:2,ls:4,rf:2,rp:1,rs:2,re:1,a:4
[4/2]aaca12lf:9,ls:3,rf:10,rs:1,a:6
[4/3]acaa12lf:8,lp:4,rf:12,rp:6,rs:1,a:4
[4/4]bcaa6lf:3,lp:3,a:8
[4/5]baaa6lf:4,lp:2,rf:1,rs:1,a:6
[4/6]aaab4lf:2,ls:2,rf:3,rs:3,a:6
[5/0]acaac8lf:4,lp:2,ls:2,rf:5,rp:2,rs:3,re:1,a:9
[5/1]caaca8lf:5,lp:3,rf:9,rp:7,rs:1,a:8
[5/2]aaaca7lf:4,ls:3,rf:1,a:6
[5/3]acaab6lf:3,lp:1,ls:2,rf:1,rp:1,rs:1,re:1,a:4
[5/4]aacab3lf:3,rf:4,rs:1,a:10
[5/5]baaaa3lf:2,lp:1,a:22
[5/6]aacaa3lf:3,rf:5,rs:1,a:8

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

Step 3

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