Morphocompletion for #2609 ⟨a, b | ababba=baba⟩

Solved by morph:2/0,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Reduction order:
- Left-to-right shortlex with a < b

Rewriting system is not complete: rulesLimit

#	Rule
1.	ababba ⇒ baba
2.	ababbbaba ⇒ babbaba
3.	ababbbbaba ⇒ babbbaba
4.	ababbbbbaba ⇒ babbbbaba
5.	ababbbabbaba ⇒ babbabbaba
6.	ababbbbbbaba ⇒ babbbbbaba
	...

Collecting factors up to length 7 / frequency 4:

[2/0]	ba	24	`lf:18,ls:6,rf:18,rp:6,rs:6,lsrp:6,lsrs:6`
[2/1]	ab	24	`lf:18,lp:6,rf:12`
[2/2]	bb	18	`lf:18,rf:12`
[3/0]	aba	22	`lf:11,lp:6,ls:5,rf:6,rs:6,lprs:6,lsrs:5`
[3/1]	bab	12	`lf:12,rf:12,rp:6`
[3/2]	bbb	11	`lf:11,rf:6,a:1`
[3/3]	bba	8	`lf:7,ls:1,rf:6`
[4/0]	abab	12	`lf:6,lp:6`
[4/1]	baba	10	`lf:5,ls:5,rf:6,rp:1,rs:6,re:1,lsrs:5`
[4/2]	babb	7	`lf:7,rf:6,rp:5`
[4/3]	bbbb	6	`lf:6,rf:3,a:3`
[5/0]	ababb	12	`lf:6,lp:6`
[5/1]	bbaba	10	`lf:5,ls:5,rf:5,rs:5,lsrs:5,a:1`
[5/2]	bbbab	5	`lf:5,rf:3,a:1`
[5/3]	babbb	5	`lf:5,rf:3,rp:3,a:1`
[6/0]	ababbb	10	`lf:5,lp:5,a:1`
[6/1]	bbbaba	8	`lf:4,ls:4,rf:3,rs:3,lsrs:3,a:1`
[6/2]	bbbbab	3	`lf:3,rf:2,a:3`
[6/3]	babbbb	3	`lf:3,rf:2,rp:2,a:3`

Considering [length 2 / frequency 0] ba=c.

Step 2

Checking up to 20 rules for overlaps.

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

Rewriting system is not complete: pendingRulesLimit

#	Rule
1.	ccbc ⇒ bcc
2.	acbc ⇒ cc
3.	ba ⇒ c
4.	ccbbcc ⇒ bcbcc
5.	acbbcc ⇒ cbcc
6.	ccbbbcc ⇒ bcbbcc
7.	ccbbcbcc ⇒ bcbcbcc
8.	acbbbcc ⇒ cbbcc
9.	acbbcbcc ⇒ cbcbcc
10.	ccbbbbcc ⇒ bcbbbcc
11.	acbbbbcc ⇒ cbbbcc
12.	acbbbcbcc ⇒ cbbcbcc
13.	acbbcbbcc ⇒ cbcbbcc
	...

Collecting factors up to length 6 / frequency 7:

[2/0]	cc	30	`lf:15,lp:5,ls:10,rf:12,rp:1,rs:12,re:1,lprs:5,lsrs:10,a:2`
[2/1]	bb	18	`lf:18,rf:8,a:4`
[2/2]	bc	18	`lf:16,ls:2,rf:19,rp:5,lsrp:1,a:2`
[2/3]	cb	16	`lf:16,rf:14,rp:6,a:2`
[2/4]	ac	14	`lf:7,lp:7,a:2`
[3/0]	bcc	20	`lf:10,ls:10,rf:11,rp:1,rs:11,re:1,lsrs:10,a:3`
[3/1]	acb	14	`lf:7,lp:7,a:2`
[3/2]	cbb	11	`lf:11,rf:6,rp:3,a:4`
[3/3]	bbc	11	`lf:11,rf:6,a:4`
[3/4]	ccb	10	`lf:5,lp:5,a:3`
[3/5]	bbb	7	`lf:7,rf:2,a:6`
[3/6]	cbc	7	`lf:5,ls:2,rf:8,rp:3,a:2`
[4/0]	bbcc	14	`lf:7,ls:7,rf:5,rs:5,lsrs:5,a:4`
[4/1]	acbb	12	`lf:6,lp:6,a:4`
[4/2]	ccbb	8	`lf:4,lp:4,a:5`
[4/3]	cbbc	6	`lf:6,rf:4,rp:2,a:4`
[4/4]	cbcc	6	`lf:3,ls:3,rf:5,rp:1,rs:5,re:1,lsrs:3,a:4`
[4/5]	cbbb	5	`lf:5,rf:2,rp:1,a:6`
[4/6]	bbbc	5	`lf:5,rf:2,a:6`
[5/0]	bbbcc	8	`lf:4,ls:4,rf:2,rs:2,lsrs:2,a:6`
[5/1]	acbbb	6	`lf:3,lp:3,a:6`
[5/2]	acbbc	6	`lf:3,lp:3,a:4`
[5/3]	cbbcc	6	`lf:3,ls:3,rf:3,rp:1,rs:3,re:1,lsrs:1,a:4`
[5/4]	bcbcc	6	`lf:3,ls:3,rf:4,rp:1,rs:4,re:1,lsrs:3,a:5`
[5/5]	ccbbb	4	`lf:2,lp:2,a:11`
[5/6]	ccbbc	4	`lf:2,lp:2,a:5`

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

Step 3

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

Rewriting system is complete. See ⟨a, b | ababba=baba⟩.