Morphocompletion for #7105 ⟨a, b, c | ab=1, cbbcc=b⟩

Solved by morph:2/1,4/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Reduction order:
- Left-to-right recursive path with deg(c) = 0; deg(a) = deg(b) = 1, a < b

Rewriting system is not complete: roundsLimit

#	Rule
1.	ab ⇒ 1
2.	cbbcc ⇒ b
3.	cbbcb ⇒ bbbcc
4.	acbbb ⇒ bbccbcc
5.	caccbbb ⇒ bbccccbcc
6.	bbbccbcc ⇒ cbbb
7.	bbccbccccbcc ⇒ accbbb
8.	aaccbbb ⇒ bccbccccbcc
9.	baccbbb ⇒ ccbbb
10.	bacccbbb ⇒ cccbbb
11.	baccccbbb ⇒ ccccbbb
12.	bacccccbbb ⇒ cccccbbb
13.	cbbbbbcc ⇒ bbbccbcb
14.	bacbcbbb ⇒ cbcbbb
15.	bacbccbbb ⇒ cbccbbb
16.	bacbcccbbb ⇒ cbcccbbb
17.	baccbcbbb ⇒ ccbcbbb
18.	baccbccbbb ⇒ ccbccbbb
19.	bacccbcbbb ⇒ cccbcbbb
20.	bacbcbcbbb ⇒ cbcbcbbb
	...

Collecting factors up to length 4 / frequency 4:

[2/0]	bb	53	`lf:37,lp:2,ls:14,rf:32,rp:4,rs:13,lsrp:2,lprs:2,lsrs:11,a:1`
[2/1]	cb	33	`lf:29,lp:3,ls:1,rf:27,rp:5,rs:1,lprs:1,a:1`
[2/2]	cc	33	`lf:29,ls:4,rf:32,rp:7,rs:4,a:1`
[2/3]	ba	22	`lf:11,lp:11,a:12`
[3/0]	bbb	33	`lf:18,lp:1,ls:14,rf:15,rp:2,rs:13,lprs:1,lsrs:11,a:2`
[3/1]	bac	22	`lf:11,lp:11,a:12`
[3/2]	cbb	20	`lf:17,lp:3,rf:13,rp:1,a:1`
[3/3]	ccb	15	`lf:15,rf:16,rp:3,a:3`

Considering [length 2 / frequency 1] cb=d.

Step 2

Checking up to 20 rules for overlaps.

Auxiliary generators:
1. d = cb
Reduction order:
- Left-to-right recursive path with deg(c) = deg(d) = 0, c < d; deg(a) = deg(b) = 1, a < b

Rewriting system is not complete: roundsLimit

#	Rule
1.	cb ⇒ d
2.	dbcc ⇒ b
3.	cdbcd ⇒ db
4.	ddbcdcc ⇒ cdbd
5.	ccdbd ⇒ ddbcdcdcccc
6.	cddbd ⇒ ddbcdcdcdcc
7.	cddbcd ⇒ ddbcdcdcc
8.	ab ⇒ 1
9.	adbd ⇒ dbcdcc
10.	adbcd ⇒ b
11.	acdbd ⇒ dbcdcdcccc
12.	addbd ⇒ dbcdcdcdcc
13.	addbcd ⇒ dbcdcdcc
14.	bb ⇒ dbcd
15.	bdbcdcc ⇒ dbd
16.	bdbcdcdcc ⇒ ddbcd
17.	bdbcdcdcdcc ⇒ ddbd
18.	dbdb ⇒ bdbcdcd
19.	dbcdb ⇒ bdbcd
20.	bdbcdcdb ⇒ dbddbcd
	...

Collecting factors up to length 6 / frequency 3:

[2/0]	db	26	`lf:20,lp:3,ls:3,rf:17,rp:8,rs:1,re:1,lsrp:1,a:3`
[2/1]	cd	26	`lf:19,lp:3,ls:4,rf:22,rp:1,rs:5,a:4`
[2/2]	bd	19	`lf:10,lp:4,ls:5,rf:6,rp:2,rs:3,lprs:2,a:7`
[3/0]	bcd	14	`lf:10,ls:4,rf:12,rs:4,a:4`
[3/1]	dbc	13	`lf:11,lp:2,rf:12,rp:5,a:3`
[3/2]	dbd	12	`lf:6,lp:1,ls:5,rf:4,rp:2,rs:3,re:1,a:8`
[4/0]	dbcd	15	`lf:10,lp:1,ls:4,rf:12,rp:5,rs:4,re:1,lsrp:1,lprs:10,a:4`
[4/1]	bdbc	8	`lf:4,lp:4,rf:2,rp:2,a:7`
[4/2]	cdcc	8	`lf:4,ls:4,rf:7,rs:5,a:8`
[5/0]	bdbcd	8	`lf:4,lp:4,rf:2,rp:2,rs:1,re:1,a:7`
[5/1]	ddbcd	6	`lf:3,lp:1,ls:2,rf:5,rp:4,rs:2,re:1,lsrp:1,a:9`
[5/2]	dbcdc	5	`lf:5,rf:8,rp:4,a:8`

Considering [length 4 / frequency 1] bdbc=e.

Step 3

Auxiliary generators:
1. d = cb
2. e = bdbc
Reduction order:
- Left-to-right recursive path with deg(c) = deg(d) = 0, c < d; deg(a) = deg(e) = deg(b) = 1, a < e < b

Rewriting system is complete. See ⟨a, b, c | ab=1, cbbcc=b⟩.