Morphocompletion for #4306 ⟨a, b | ababbaaab=aa⟩

Solved by morph:5/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aaabbaaab ⇒ ababbaaaa
2. ababbaaab ⇒ aa
3. aaaaabaabbaaab ⇒ aaaaabbabbaaaa
4. aaabbababbaaaa ⇒ ababbaaaabaaab
5. aaaaabbabbaaab ⇒ aaaaaba
6. ababbababbaaaa ⇒ aabaaab
7. aabaaabbabbaaab ⇒ aabaaaba
8. aaaaababbabbaaab ⇒ aaaaababa
9. aaaaabaabbabbaaab ⇒ aaaaabaaba
10. aabaaababbabbaaab ⇒ aabaaababa
11. aabaaabaabbabbaaab ⇒ aabaaabaaba
...

Collecting factors up to length 8 / frequency 7:

[2/0]aa67lf:56,lp:9,ls:2,rf:42,rp:9,rs:3,re:1,lprp:7,lsrp:1,lprs:2
[2/1]ab51lf:40,lp:2,ls:9,rf:23,rp:2,rs:2,lsrp:1,lprs:1
[2/2]ba31lf:31,rf:21,rs:6
[2/3]bb19lf:19,rf:4
[3/0]aaa38lf:30,lp:6,ls:2,rf:23,rp:4,rs:2,lprp:4,lprs:2
[3/1]aab36lf:24,lp:3,ls:9,rf:16,rp:4,rs:2,lprp:3,lsrp:3
[3/2]abb19lf:19,rf:4
[3/3]bba19lf:19,rf:4
[3/4]baa17lf:17,rf:10
[3/5]bab14lf:14,rf:5
[3/6]aba14lf:12,lp:2,rf:17,rp:2,rs:6
[4/0]aaab29lf:18,lp:2,ls:9,rf:10,rs:2,lprs:1
[4/1]abba19lf:19,rf:4
[4/2]aaaa16lf:10,lp:4,ls:2,rf:11,rp:4,rs:2,lprp:4,lprs:1,a:1
[4/3]baaa14lf:14,rf:8
[4/4]babb12lf:12,rf:3
[4/5]bbaa11lf:11,rf:3
[4/6]aaba11lf:8,lp:3,rf:13,rp:4,rs:4,lprp:3,lprs:2,a:2
[5/0]baaab21lf:12,ls:9,rf:5,rs:2
[5/1]babba12lf:12,rf:3
[5/2]abbaa11lf:11,rf:3
[5/3]bbaaa11lf:11,rf:3
[5/4]aabaa9lf:6,lp:3,rf:7,rp:4,lprp:3,a:2
[5/5]ababb8lf:6,lp:2,rf:2,rp:2
[5/6]abbab8lf:8,rf:1,a:2
[6/0]bbaaab18lf:9,ls:9
[6/1]abbaaa11lf:11,rf:3
[6/2]babbaa9lf:9,rf:3
[6/3]ababba8lf:6,lp:2,rf:2,rp:2
[6/4]aaaaab8lf:4,lp:4,rf:4,rp:4,lprp:4,a:8
[6/5]abbabb6lf:6,rf:1,a:4
[6/6]bbabba6lf:6,rf:1,a:4
[7/0]abbaaab18lf:9,ls:9
[7/1]babbaaa9lf:9,rf:3
[7/2]abbabba6lf:6,rf:1,a:4
[7/3]bbabbaa6lf:6,rf:1,a:4
[7/4]aabaaab6lf:3,lp:3,rf:5,rp:4,rs:2,re:1,lprp:3,a:2
[7/5]aaaaaba6lf:3,lp:3,rf:3,rp:3,rs:1,re:1,lprp:2,a:8
[7/6]aabbabb4lf:4,rf:1,a:4

Considering [length 5 / frequency 1] babba=c.

Step 2

Rewriting system is complete. See ⟨a, b | ababbaaab=aa⟩.