Morphocompletion for #2918 ⟨a, b | aaabaababba=1⟩

Solved by morph:6/0,4/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Reduction order:
- Right-to-left recursive path with deg(a) = 0; deg(b) = 1

Rewriting system is not complete: arenaTotalLimit

#	Rule
1.	baaaabaaba ⇒ aaaabaabab
2.	babbaaaa ⇒ aabaabab
3.	aaaabaababb ⇒ 1
4.	baaaabaabab ⇒ 1
5.	baababba ⇒ abaababb
6.	aabaabababaababb ⇒ babba
7.	aabaababaabaababb ⇒ babbaa
8.	aabaababaaabaababb ⇒ babbaaa
9.	aabaababbbabba ⇒ baabababaababb
10.	aabaababbbaababbbabba ⇒ baababbbaabababaababb
11.	aabaababbbaababbbaababbbabba ⇒ baababbbaababbbaabababaababb
12.	aabaababbbaababbbaababbbaababbbabba ⇒ baababbbaababbbaababbbaabababaababb
13.	aabaababbbaababbbaababbbaababbbaababbbabba ⇒ baababbbaababbbaababbbaababbbaabababaababb
14.	aabaababbbaababbbaababbbaababbbaababbbaababbbabba ⇒ baababbbaababbbaababbbaababbbaababbbaabababaababb
15.	aabaababbbaababbbaababbbaababbbaababbbaababbbaababbbabba ⇒ baababbbaababbbaababbbaababbbaababbbaababbbaabababaababb
	...

Collecting factors up to length 7 / frequency 4:

[2/0]	ba	111	`lf:98,lp:4,ls:9,rf:89,rp:10,rs:1,lsrp:7,a:3`
[2/1]	ab	100	`lf:99,ls:1,rf:89,rp:1,rs:2,a:3`
[2/2]	aa	75	`lf:63,lp:11,ls:1,rf:45,rp:2,rs:2,lsrp:1,lprs:2,a:3`
[2/3]	bb	73	`lf:69,ls:4,rf:53,rs:8,a:3`
[3/0]	aab	63	`lf:53,lp:10,rf:40,rp:1,a:3`
[3/1]	aba	58	`lf:57,ls:1,rf:55,rp:1,a:3`
[3/2]	bab	48	`lf:46,lp:1,ls:1,rf:48,rp:3,rs:2,lprs:1,a:3`
[3/3]	baa	46	`lf:43,lp:3,rf:40,rp:7,rs:1,a:3`
[4/0]	aaba	64	`lf:53,lp:10,ls:1,rf:40,rp:1,a:3`
[4/1]	babb	46	`lf:41,lp:1,ls:4,rf:32,rp:3,rs:8,lsrp:3,a:3`
[4/2]	baab	40	`lf:39,lp:1,rf:38,rp:7,a:3`
[4/3]	abab	39	`lf:38,ls:1,rf:45,rs:2,a:3`
[5/0]	baaba	41	`lf:39,lp:1,ls:1,rf:38,rp:7,a:3`
[5/1]	aabab	38	`lf:37,ls:1,rf:38,rs:2,a:3`
[5/2]	ababb	37	`lf:33,ls:4,rf:29,rs:8,a:3`
[5/3]	babbb	28	`lf:28,rf:21,a:4`
[6/0]	baabab	39	`lf:37,lp:1,ls:1,rf:38,rp:7,rs:2,a:3`
[6/1]	aababb	37	`lf:33,ls:4,rf:29,rs:8,a:3`
[6/2]	babbba	28	`lf:28,rf:21,a:4`
[6/3]	ababbb	28	`lf:28,rf:21,a:4`

Considering [length 6 / frequency 0] baabab=c.

Step 2

Checking up to 20 rules for overlaps.

Auxiliary generators:
1. c = baabab
Reduction order:
- Right-to-left recursive path with deg(a) = 0; deg(b) = deg(c) = 1, b < c

Rewriting system is not complete: roundsLimit

#	Rule
1.	aaaacb ⇒ 1
2.	cba ⇒ acb
3.	baaaac ⇒ 1
4.	aacc ⇒ bab
5.	aaacaac ⇒ bbaaaa
6.	caaaac ⇒ baaba
7.	baabab ⇒ c
8.	babbaaaa ⇒ aac
9.	aacacb ⇒ babba
10.	aacaacb ⇒ babbaa
11.	aacaaacb ⇒ babbaaa
12.	baabac ⇒ caabab
13.	aacbbab ⇒ cc
14.	acbbbaaaa ⇒ caac
15.	aacbbac ⇒ ccaabab
16.	aacbcc ⇒ cbbab
17.	aacbcbbab ⇒ cbcc
	...

Collecting factors up to length 6 / frequency 7:

[2/0]	aa	43	`lf:31,lp:10,ls:2,rf:11,rp:1,rs:3,lsrp:1,lprs:3,a:1`
[2/1]	ac	25	`lf:19,lp:1,ls:5,rf:3,rp:1,rs:2,lprs:1,a:4`
[2/2]	ba	17	`lf:12,lp:4,ls:1,rf:13,rp:5,rs:2,a:1`
[2/3]	cb	16	`lf:11,lp:1,ls:4,rf:3,rp:2,rs:1,lprs:10,a:9`
[2/4]	ab	9	`lf:6,ls:3,rf:10,rs:4,a:1`
[2/5]	bb	6	`lf:6,rf:5,rp:1,a:10`
[2/6]	ca	6	`lf:5,lp:1,rf:3,rp:2,a:4`
[3/0]	aac	26	`lf:15,lp:8,ls:3,rf:2,rp:1,rs:2,re:1,a:6`
[3/1]	aaa	16	`lf:12,lp:2,ls:2,rf:3,rs:2,lprs:1,a:21`
[3/2]	acb	14	`lf:9,lp:1,ls:4,rf:1,rp:1,rs:1,re:1,a:9`
[3/3]	bab	8	`lf:4,lp:1,ls:3,rf:7,rp:4,rs:4,re:1,a:1`
[3/4]	baa	8	`lf:5,lp:3,rf:4,rp:1,rs:1,a:1`
[3/5]	bba	5	`lf:5,rf:5,rp:1,rs:1,a:10`
[3/6]	caa	5	`lf:4,lp:1,rf:3,rp:2,a:4`
[4/0]	aacb	14	`lf:7,lp:4,ls:3,a:10`
[4/1]	aaac	8	`lf:5,lp:1,ls:2,a:21`
[4/2]	aaaa	8	`lf:5,lp:1,ls:2,rf:1,rs:1,a:21`
[4/3]	aaca	7	`lf:4,lp:3,a:12`
[4/4]	acbb	4	`lf:3,lp:1,a:10`
[4/5]	bbab	4	`lf:2,ls:2,rf:1,rs:1,a:10`
[4/6]	baab	4	`lf:2,lp:2,rf:1,rp:1,a:1`
[5/0]	aaaac	6	`lf:3,lp:1,ls:2,a:21`
[5/1]	baaaa	6	`lf:3,lp:1,ls:2,rf:1,rs:1,a:28`
[5/2]	aacaa	5	`lf:3,lp:2,a:18`
[5/3]	cbbab	4	`lf:2,ls:2,rf:1,rp:1,rs:1,re:1,a:10`
[5/4]	aacbc	4	`lf:2,lp:2,a:14`
[5/5]	aacbb	4	`lf:2,lp:2,a:10`
[5/6]	aaacb	4	`lf:2,ls:2,a:21`

Considering [length 4 / frequency 0] aacb=d.

Step 3

Auxiliary generators:
1. c = baabab
2. d = aacb
Reduction order:
- Right-to-left recursive path with deg(a) = deg(d) = 0, a < d; deg(b) = deg(c) = 1, b < c

Rewriting system is complete. See ⟨a, b | aaabaababba=1⟩.