Morphocompletion for #6222 ⟨a, b, c | aa=1, bacacb=1⟩

Back:⟨a, b, c | aa=1, bacacb=1⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. cacbb ⇒ a
2. bbac ⇒ acbb
3. aa ⇒ 1
4. bacac ⇒ acacb
5. cacbacbb ⇒ abac
6. cacbacbacbb ⇒ abacbac
7. cacbacbacbacbb ⇒ abacbacbac
8. cacbacbacbacbacbb ⇒ abacbacbacbac
9. cacbacbacbacbacbacbb ⇒ abacbacbacbacbac
10. cacbacbacbacbacbacbacbb ⇒ abacbacbacbacbacbac
11. cacbacbacbacbacbacbacbacbb ⇒ abacbacbacbacbacbacbac
12. cacbacbacbacbacbacbacbacbacbb ⇒ abacbacbacbacbacbacbacbac
13. cacbacbacbacbacbacbacbacbacbacbb ⇒ abacbacbacbacbacbacbacbacbac
14. cacbacbacbacbacbacbacbacbacbacbacbb ⇒ abacbacbacbacbacbacbacbacbacbac
15. cacbacbacbacbacbacbacbacbacbacbacbacbb ⇒ abacbacbacbacbacbacbacbacbacbacbac
16. cacbacbacbacbacbacbacbacbacbacbacbacbacbb ⇒ abacbacbacbacbacbacbacbacbacbacbacbac
17. cacbacbacbacbacbacbacbacbacbacbacbacbacbacbb ⇒ abacbacbacbacbacbacbacbacbacbacbacbacbac
...

Collecting factors up to length 4 / frequency 7:

[2/0]ac110lf:108,ls:2,rf:94,rp:2,rs:13,lsrp:2,a:2
[2/1]cb105lf:105,rf:80,rs:1,a:2
[2/2]ba94lf:93,lp:1,rf:91,a:2
[2/3]bb30lf:15,lp:1,ls:14,rf:1,rs:1,lprs:1,a:4
[2/4]ca29lf:15,lp:14,rf:1,a:2
[2/5]ab0rf:13,rp:13,a:6
[3/0]acb105lf:105,rf:80,rp:1,rs:1,a:2
[3/1]bac95lf:93,lp:1,ls:1,rf:91,rs:13,a:2
[3/2]cba91lf:91,rf:78,a:6
[3/3]cac30lf:15,lp:14,ls:1,rf:1,a:2
[3/4]cbb28lf:14,ls:14,rf:1,rs:1,a:4
[3/5]bba2lf:1,lp:1,a:7
[3/6]aca1lf:1,rf:1,rp:1,a:2

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

Step 2

Rewriting system is complete. See ⟨a, b, c | aa=1, bacacb=1⟩.