Morphocompletion for #568 ⟨a, b | abab=aaba⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abab ⇒ aaba
2. abaaba ⇒ aabaab
3. abaaaba ⇒ aabaabb
4. abaaaaba ⇒ aabaabbb
5. aabaabbab ⇒ aaabaabba
6. abaaaaaba ⇒ aabaabbbb
7. aabaabbbab ⇒ aaabaabbaa
8. abaaaaaaba ⇒ aabaabbbbb
9. aabaabbaaba ⇒ aaabaabbaab
10. aabaabbbbab ⇒ aaabaabbaaa
11. abaaaaaaaba ⇒ aabaabbbbbb
...

Collecting factors up to length 8 / frequency 7:

[2/0]ab37lf:26,lp:7,ls:4,rf:22,rs:2,lprs:1
[2/1]aa34lf:30,lp:4,rf:29,rp:11,rs:2,lprp:4,lprs:2
[2/2]ba29lf:22,ls:7,rf:15,rs:2
[2/3]bb7lf:7,rf:19,rs:5,a:3
[3/0]aba32lf:18,lp:7,ls:7,rf:11,rs:1,lprs:10
[3/1]aab19lf:15,lp:4,rf:22,rp:7,rs:2,lprs:1
[3/2]aaa15lf:15,rf:5,rp:4,rs:1,a:3
[3/3]baa11lf:11,rf:13,rs:1,a:1
[3/4]bab8lf:4,ls:4
[3/5]bba4lf:4,rf:4,rs:1,a:4
[3/6]abb4lf:4,rf:9,rs:1,a:3
[4/0]aaba22lf:11,lp:4,ls:7,rf:11,rp:7,rs:1,re:1,lsrp:6
[4/1]abaa16lf:10,lp:6,rf:10,a:1
[4/2]aaaa10lf:10,a:6
[4/3]bbab6lf:3,ls:3,a:4
[4/4]baab6lf:6,rf:11,rs:2,a:1
[4/5]baaa5lf:5,rf:1,rs:1,a:3
[4/6]aaab5lf:5,rf:4,rp:4,a:3
[5/0]abaaa10lf:5,lp:5,a:3
[5/1]aaaba10lf:5,ls:5,rf:4,rp:4,a:3
[5/2]aabaa8lf:4,lp:4,rf:10,rp:6,a:1
[5/3]abaab6lf:5,lp:1,rf:10,rs:1,lprs:10,a:1
[5/4]aaaaa6lf:6,a:11
[5/5]bbbab4lf:2,ls:2,a:7
[5/6]baabb4lf:4,rf:9,rs:1,a:3
[6/0]abaaaa8lf:4,lp:4,a:6
[6/1]aabaab8lf:4,lp:4,rf:10,rp:6,rs:1,re:1,a:1
[6/2]aaaaba8lf:4,ls:4,a:6
[6/3]abaabb4lf:4,rf:9,rs:1,a:3
[6/4]baaaaa3lf:3,a:11
[6/5]aaaaab3lf:3,a:11
[6/6]aaaaaa3lf:3,a:21
[7/0]aabaabb8lf:4,lp:4,rf:9,rp:5,rs:1,re:1,a:3
[7/1]abaaaaa6lf:3,lp:3,a:11
[7/2]aaaaaba6lf:3,ls:3,a:11
[7/3]baabbab2lf:1,ls:1,a:4
[7/4]baaaaba2lf:1,ls:1,a:6
[7/5]baaaaaa2lf:2,a:21
[7/6]abbbbab2lf:1,ls:1,a:12

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

Step 2

Rewriting system is complete. See ⟨a, b | abab=aaba⟩.