Morphocompletion for #4303 ⟨a, b | abababbba=ab

Solved by morph:2/1,2/0,4/1. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abababb ⇒ abbbbaa
2. abbbbaab ⇒ abbabbba
3. abbababb ⇒ abbbbbaa
4. abbabbbaa ⇒ ab
5. abbbbbaab ⇒ abbbabbba
6. abbbababb ⇒ abbbbbbaa
7. abbbabbbaa ⇒ abb
8. abbbbabbbaa ⇒ abbb
9. abbbbbabbbaa ⇒ abbbb
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] bb, [2/1] ab, [2/2] ba
Length 3:[3/0] abb, [3/1] bbb, [3/2] bba
Length 4:[4/0] abbb, [4/1] babb, [4/2] bbba
Length 5:[5/0] bbbaa, [5/1] abbbb, [5/2] abbba
Length 6:[6/0] abbbaa, [6/1] bababb, [6/2] abbbbb

Considering [length 2 / frequency 1] ab=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ab ⇒ c
2. cccbbc ⇒ cb
3. cccbba ⇒ c
4. cbcbbc ⇒ cccbbb
5. cbccbbc ⇒ cbb
6. cbbba ⇒ cccbccbb
7. cbcbba ⇒ cccbb
8. cbccbba ⇒ cb
9. cccbccbbb ⇒ cbbbc
10. cbbcbbc ⇒ cbccbbb
11. cbbccbbc ⇒ cbbb
12. cbbbba ⇒ cbccbccbb
13. cbbcbba ⇒ cbccbb
14. cbbccbba ⇒ cbb
15. cbccbccbbb ⇒ cbbbbc
16. cbbbcbbc ⇒ cbbccbbb
17. cbbbccbbc ⇒ cbbbb
18. cbbbcbba ⇒ cbbccbb
19. cbbbccbba ⇒ cbbb
20. cbbbbccbba ⇒ cbbbb
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] cb, [2/1] bb, [2/2] bc, [2/3] ba, [2/4] cc
Length 3:[3/0] cbb, [3/1] bbc, [3/2] bba, [3/3] bbb, [3/4] ccb
Length 4:[4/0] cbbc, [4/1] cbbb, [4/2] cbba, [4/3] ccbb, [4/4] bccb
Length 5:[5/0] ccbba, [5/1] bccbb, [5/2] cbbbc, [5/3] ccbbc, [5/4] cbccb

Considering [length 2 / frequency 0] cb=d.

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. cb ⇒ d
2. ccdbd ⇒ db
3. ddbc ⇒ ccdcdbd
4. dcdbc ⇒ db
5. ccdbc ⇒ d
6. dcdbdc ⇒ ccdcdcdbd
7. ddba ⇒ ccdb
8. dcdba ⇒ d
9. ccdba ⇒ c
10. ddbda ⇒ ccdcdcdb
11. dcdbda ⇒ ccdcdb
12. ab ⇒ c
13. dbb ⇒ dcdbd
14. ccdcdbdb ⇒ ddbd
15. dbdbd ⇒ dcdcdbdb
16. dbcdbd ⇒ dcdbdb
17. dbdbc ⇒ dcdcdbd
18. dbcdbc ⇒ dcdbd
19. dbdba ⇒ dcdb
20. dbcdba ⇒ db
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] db, [2/1] bc, [2/2] bd, [2/3] cd, [2/4] dc
Length 3:[3/0] dbc, [3/1] dbd, [3/2] cdb, [3/3] dba, [3/4] dcd
Length 4:[4/0] dcdb, [4/1] dbdb, [4/2] cdbd, [4/3] cdba, [4/4] dbcd
Length 5:[5/0] dbcdb, [5/1] dcdbd, [5/2] bcdba, [5/3] bcdbc, [5/4] bcdbd

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

Step 4

Rewriting system is complete. See a, b | abababbba=ab.