Morphocompletion for #3761 ⟨a, b | ababaabaab=a

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. ababa ⇒ aaabb
2. aaabbba ⇒ abaaabb
3. aaabbaba ⇒ aaabaabb
4. aaabaabba ⇒ aabaabaab
5. aabaabaabb ⇒ a
6. aaabaabbba ⇒ aaabbaaabb
7. aaababbaba ⇒ aaaaabbabb
8. aaabbaabbba ⇒ abaaabbaabb
9. aaaaabbabba ⇒ aaabbaabaab
10. aaabbaabbaba ⇒ aaabbaabaabb
11. aaabbaabaabba ⇒ aaab
12. aabaabaaaabbabba ⇒ aaabaab
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] aa, [2/1] ba, [2/2] ab
Length 3:[3/0] aab, [3/1] aaa, [3/2] bba
Length 4:[4/0] aaab, [4/1] abba, [4/2] aabb
Length 5:[5/0] aaabb, [5/1] aabba, [5/2] aabaa
Length 6:[6/0] aaabba, [6/1] aabaab, [6/2] abbaba

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

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ca ⇒ ac
2. aa ⇒ c
3. ccbba ⇒ cbabc
4. acbba ⇒ ababc
5. cbaba ⇒ ccbb
6. ababa ⇒ acbb
7. ccbbba ⇒ cbacbb
8. acbbba ⇒ abacbb
9. cbcbcbb ⇒ a
10. cbabcbc ⇒ acbcbcb
11. ababcbc ⇒ cbcbcb
12. ccbcbba ⇒ acbcbcb
13. acbcbba ⇒ cbcbcb
14. cbcbcba ⇒ acbcbbc
15. cbabcba ⇒ ccbcbb
16. ababcba ⇒ acbcbb
17. ccbbcbcbba ⇒ ccb
18. acbbcbcbba ⇒ acb
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] ba, [2/1] cb, [2/2] bc, [2/3] bb, [2/4] ab
Length 3:[3/0] bba, [3/1] cbc, [3/2] bcb, [3/3] cba, [3/4] cbb
Length 4:[4/0] cbba, [4/1] cbcb, [4/2] bcbc, [4/3] bcba, [4/4] abab
Length 5:[5/0] bcbba, [5/1] cbcbb, [5/2] abcba, [5/3] cbbba, [5/4] ababc

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

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. bc ⇒ da
2. dddddadc ⇒ dcd
3. ba ⇒ d
4. ca ⇒ ac
5. aa ⇒ c
6. dddddada ⇒ dcb
7. dddadad ⇒ da
8. dcbb ⇒ ddd
9. dddadab ⇒ d
10. cddddadc ⇒ ccd
11. dddddadac ⇒ dcda
12. addddadc ⇒ acd
13. cdadad ⇒ addadc
14. cddadad ⇒ ac
15. dadadad ⇒ dddadc
16. addadad ⇒ c
17. cdadab ⇒ addada
18. cddadab ⇒ c
19. dadadab ⇒ dddada
20. addadab ⇒ a
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] ad, [2/1] da, [2/2] dd, [2/3] ab, [2/4] cd
Length 3:[3/0] dad, [3/1] ddd, [3/2] ada, [3/3] dda, [3/4] dab
Length 4:[4/0] dada, [4/1] adad, [4/2] ddad, [4/3] dddd, [4/4] adab
Length 5:[5/0] dadad, [5/1] dadab, [5/2] ddada, [5/3] dddad, [5/4] ddadc

Considering [length 3 / frequency 0] dad=e.

Step 4

Rewriting system is complete. See a, b | ababaabaab=a.