Morphocompletion for #2265 ⟨a, b | aabbaba=baa⟩

Solved by morph:4/1,3/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

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

Rewriting system is not complete: rulesLimit

#	Rule
1.	aabbaba ⇒ baa
2.	aabbabbaa ⇒ babaa
3.	aabbabbbaa ⇒ babbaa
4.	aabbabbbbaa ⇒ babbbaa
5.	aabbabbbbbaa ⇒ babbbbaa
6.	aabbabbabaa ⇒ bababaa
	...

Collecting factors up to length 7 / frequency 4:

[2/0]	aa	22	`lf:11,lp:6,ls:5,rf:6,rs:6,lprs:6,lsrs:5`
[2/1]	bb	17	`lf:17,rf:6`
[2/2]	ba	14	`lf:13,ls:1,rf:12,rp:6,lsrp:1`
[2/3]	ab	13	`lf:13,rf:6`
[3/0]	aab	12	`lf:6,lp:6`
[3/1]	abb	11	`lf:11,rf:3`
[3/2]	bba	11	`lf:11,rf:3`
[3/3]	baa	10	`lf:5,ls:5,rf:6,rp:1,rs:6,re:1,lsrs:5`
[4/0]	aabb	12	`lf:6,lp:6`
[4/1]	abba	8	`lf:8,rf:1`
[4/2]	bbaa	8	`lf:4,ls:4,rf:3,rs:3,lsrs:3,a:1`
[4/3]	bbab	7	`lf:7`
[5/0]	aabba	12	`lf:6,lp:6`
[5/1]	abbab	7	`lf:7`
[5/2]	bbbaa	6	`lf:3,ls:3,rf:2,rs:2,lsrs:2,a:3`
[5/3]	bbabb	5	`lf:5,a:1`
[6/0]	aabbab	12	`lf:6,lp:6`
[6/1]	abbabb	5	`lf:5,a:1`
[6/2]	bbbbaa	4	`lf:2,ls:2,rf:1,rs:1,lsrs:1,a:9`
[6/3]	abbaba	3	`lf:2,ls:1`

Considering [length 4 / frequency 1] abba=c.

Step 2

Checking up to 20 rules for overlaps.

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

Rewriting system is not complete: pendingRulesLimit

#	Rule
1.	acbc ⇒ bac
2.	abbc ⇒ cbba
3.	acbbcc ⇒ babcc
4.	abbbcc ⇒ ccbbbac
5.	acbbbcc ⇒ bcbbac
6.	acba ⇒ baa
7.	abba ⇒ c
8.	acbbca ⇒ babca
9.	abbbca ⇒ ccbbbaa
10.	acbbbca ⇒ bcbbaa
11.	acbbac ⇒ babac
12.	abbbac ⇒ ccbc
13.	acbbbac ⇒ bcc
14.	acbbaa ⇒ babaa
15.	abbbaa ⇒ ccba
16.	acbbbaa ⇒ bca
	...

Collecting factors up to length 6 / frequency 7:

[2/0]	ac	26	`lf:13,lp:10,ls:3,rf:4,rs:4,lprs:3,lsrs:1,a:2`
[2/1]	bb	22	`lf:22,rf:7,a:1`
[2/2]	ab	12	`lf:6,lp:6,rf:4,a:1`
[2/3]	ba	10	`lf:8,ls:2,rf:14,rp:6,rs:2,lsrp:1,a:1`
[2/4]	bc	10	`lf:8,ls:2,rf:7,rp:4,rs:1,a:3`
[2/5]	cb	10	`lf:10,rf:7,rp:1,a:2`
[2/6]	aa	6	`lf:3,ls:3,rf:4,rs:4,lsrs:1,a:2`
[3/0]	acb	20	`lf:10,lp:10,a:2`
[3/1]	abb	12	`lf:6,lp:6,a:1`
[3/2]	bba	8	`lf:7,ls:1,rf:5,rs:1,a:1`
[3/3]	bbb	8	`lf:8,rf:2,a:4`
[3/4]	bbc	8	`lf:7,ls:1,a:3`
[3/5]	cbb	8	`lf:8,rf:5,rp:1,a:3`
[3/6]	baa	6	`lf:3,ls:3,rf:4,rp:1,rs:4,re:1,lsrs:1,a:2`
[4/0]	acbb	16	`lf:8,lp:8,a:6`
[4/1]	abbb	8	`lf:4,lp:4,a:4`
[4/2]	bbaa	6	`lf:3,ls:3,rf:2,rs:2,a:4`
[4/3]	bbac	6	`lf:3,ls:3,rf:2,rs:2,a:7`
[4/4]	bbca	6	`lf:3,ls:3,a:28`
[4/5]	bbcc	6	`lf:3,ls:3,a:19`
[4/6]	bbba	4	`lf:4,rf:2,a:4`
[5/0]	acbbb	8	`lf:4,lp:4,a:16`
[5/1]	abbba	4	`lf:2,lp:2,a:4`
[5/2]	acbba	4	`lf:2,lp:2,a:6`
[5/3]	bbbaa	4	`lf:2,ls:2,rf:1,rs:1,a:4`
[5/4]	abbbc	4	`lf:2,lp:2,a:19`
[5/5]	acbbc	4	`lf:2,lp:2,a:20`
[5/6]	bbbac	4	`lf:2,ls:2,rf:1,rs:1,a:7`

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

Step 3

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

Rewriting system is complete. See ⟨a, b | aabbaba=baa⟩.