Morphocompletion for #846 ⟨a, b | ababaaab=a

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. aaaabba ⇒ aabaaab
2. ababaa ⇒ aaaabb
3. aabaaabb ⇒ a
4. aaaabaabba ⇒ aabaaabaab
5. aabaaababba ⇒ aaaabbbaaab
6. aabaaabaabba ⇒ aaaab
7. aaaabbbaba ⇒ aaaababbab
8. aaaababbaba ⇒ aabaaabaabb
9. aaaabbbaaabb ⇒ ababa
10. aabaaabaabaabba ⇒ aaaabaab
11. aaaababbaaabba ⇒ aaaabab
12. aaaababbbaaabba ⇒ aaaababb
...

Collecting factors up to length 7, frequency 3:

Length 2:[2/0] aa, [2/1] ba, [2/2] ab
Length 3:[3/0] aaa, [3/1] aab, [3/2] bba
Length 4:[4/0] abba, [4/1] aaba, [4/2] aaab
Length 5:[5/0] aaaab, [5/1] aabba, [5/2] aabaa
Length 6:[6/0] aabaaa, [6/1] aaaaba, [6/2] baabba

Considering [length 3 / frequency 0] aaa=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ca ⇒ ac
2. aaa ⇒ c
3. cbabc ⇒ acbcb
4. ababc ⇒ aabcb
5. ccbba ⇒ acbcb
6. acbba ⇒ aabcb
7. cbabac ⇒ acbcba
8. ababac ⇒ aabcba
9. cbabaa ⇒ ccbb
10. ababaa ⇒ acbb
11. cbcbb ⇒ aa
12. aabcbb ⇒ a
13. ccbbbabc ⇒ ccb
14. acbbbabc ⇒ acb
15. ccbbcbba ⇒ ccb
16. acbbcbba ⇒ acb
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] ba, [2/1] cb, [2/2] ab, [2/3] bb, [2/4] bc
Length 3:[3/0] cbb, [3/1] bba, [3/2] aba, [3/3] abc, [3/4] bab
Length 4:[4/0] cbba, [4/1] babc, [4/2] bcbb, [4/3] abab, [4/4] cbab
Length 5:[5/0] bcbba, [5/1] bbabc, [5/2] babaa, [5/3] babac, [5/4] ababa

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

Step 3

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. dddaddc ⇒ dcd
2. ca ⇒ ac
3. dddaad ⇒ da
4. cddaddc ⇒ ccd
5. addaddc ⇒ acd
6. cdaad ⇒ aaddc
7. cddaad ⇒ ac
8. aaa ⇒ c
9. dadaad ⇒ dddc
10. addaad ⇒ aa
11. aadaad ⇒ addc
12. dcb ⇒ dddaddaa
13. dddaab ⇒ d
14. cdaab ⇒ aaddaa
15. cddaab ⇒ c
16. dadaab ⇒ dddaa
17. addaab ⇒ a
18. aadaab ⇒ addaa
19. bc ⇒ daa
20. ba ⇒ d
...

Collecting factors up to length 6, frequency 5:

Length 2:[2/0] ad, [2/1] aa, [2/2] da, [2/3] dd, [2/4] ab
Length 3:[3/0] aad, [3/1] aab, [3/2] daa, [3/3] add, [3/4] dda
Length 4:[4/0] daab, [4/1] daad, [4/2] adda, [4/3] ddaa, [4/4] cdda
Length 5:[5/0] ddaab, [5/1] ddaad, [5/2] daddc, [5/3] adaab, [5/4] aadaa

Considering [length 2 / frequency 0] ad=e.

Step 4

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