Morphocompletion for #4192 ⟨a, b | aabbbaaab=ba⟩

Back:⟨a, b | aabbbaaab=ba⟩

Solved by morph:5/0,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. aabbbaaab ⇒ ba
2. aabbbaba ⇒ babbaaab
3. babbaaababbbaaab ⇒ aabbbabba
4. aabbbaaaaabbbabba ⇒ baabbaaababbbaaab
5. babbaaababbbaba ⇒ aabbbabbabbaaab
6. baabbaaababbbaaababbbaaab ⇒ aabbbaaaaabbbabbba
7. aabbbaaaaabbbaaaaabbbabbba ⇒ baaabbaaababbbaaababbbaaab
8. babbaaababbbaaaaabbbabba ⇒ aabbbabbaabbaaababbbaaab
9. baabbaaababbbaaababbbaba ⇒ aabbbaaaaabbbabbbabbaaab
10. baabbaaababbbaaabaababbbaaab ⇒ aabbbaaaaabbaabbbabba
11. aabbbaaaaabbbaaaaabbaabbbabba ⇒ baaabbaaababbbaaabaababbbaaab
12. baabbaaababbbaaabaababbbaba ⇒ aabbbaaaaabbaabbbabbabbaaab
...

Collecting factors up to length 7 / frequency 4:

[2/0]ba73lf:58,lp:7,ls:8,rf:46,rp:5,rs:4,re:1,lsrp:4,lprs:3
[2/1]aa71lf:66,lp:5,rf:60,rp:7
[2/2]ab59lf:55,ls:4,rf:49,rs:8
[2/3]bb57lf:57,rf:53
[3/0]aab42lf:33,lp:5,ls:4,rf:32,rp:7,rs:8,lsrp:3,lprs:4
[3/1]bba38lf:34,ls:4,rf:34,rs:3
[3/2]abb34lf:34,rf:34
[3/3]aaa33lf:33,rf:28
[4/0]aaab25lf:21,ls:4,rf:20,rs:8
[4/1]bbba24lf:23,ls:1,rf:19,rs:1
[4/2]abbb23lf:23,rf:19
[4/3]bbaa22lf:22,rf:21
[5/0]abbba24lf:23,ls:1,rf:19,rs:1
[5/1]bbaaa21lf:21,rf:18
[5/2]baaab19lf:15,ls:4,rf:16,rp:2,rs:8
[5/3]aabbb16lf:11,lp:5,rf:11,rp:7
[6/0]bbaaab19lf:15,ls:4,rf:14,rs:8
[6/1]aabbba16lf:11,lp:5,rf:11,rp:7
[6/2]bbbaaa14lf:14,rf:10
[6/3]abbbaa14lf:14,rf:10

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: lhsLimit

#Rule
1. ba ⇒ acaab
2. aacaabcaabcaab ⇒ c
...

Collecting factors up to length 6 / frequency 7:

[2/0]aa5lf:4,lp:1,rf:1,a:2
[2/1]ab4lf:3,ls:1,rf:1,rs:1,a:2
[2/2]ca3lf:3,rf:1,a:2
[2/3]bc2lf:2,a:3
[2/4]ac1lf:1,rf:1,rp:1,a:2
[3/0]aab4lf:3,ls:1,rf:1,rs:1,a:2
[3/1]caa3lf:3,rf:1,a:2
[3/2]bca2lf:2,a:3
[3/3]abc2lf:2,a:3
[3/4]aac2lf:1,lp:1,a:3
[3/5]aca1lf:1,rf:1,rp:1,a:2
[4/0]caab4lf:3,ls:1,rf:1,rs:1,a:2
[4/1]bcaa2lf:2,a:3
[4/2]abca2lf:2,a:3
[4/3]aabc2lf:2,a:3
[4/4]aaca2lf:1,lp:1,a:3
[4/5]acaa1lf:1,rf:1,rp:1,a:2
[5/0]bcaab3lf:2,ls:1,a:3
[5/1]abcaa2lf:2,a:3
[5/2]aabca2lf:2,a:3
[5/3]caabc2lf:2,a:3
[5/4]aacaa2lf:1,lp:1,a:3
[5/5]acaab1lf:1,rf:1,rp:1,rs:1,re:1,a:2

Considering [length 3 / frequency 0] aab=d.

Step 3

Rewriting system is complete. See ⟨a, b | aabbbaaab=ba⟩.