Morphocompletion for #1803 ⟨a, b | abbaabaab=a⟩

Back:⟨a, b | abbaabaab=a⟩

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

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: rulesLimit

#Rule
1. abbaa ⇒ aaabb
2. aaabaabbb ⇒ a
3. aaabbbaa ⇒ aaabaabb
4. aaabbbbaa ⇒ aaaaabbbb
5. abaabaa ⇒ aaabaab
6. aaabbabaabbb ⇒ abba
7. aaaaabbbbbaa ⇒ aaab
8. aaabaaaabbbb ⇒ abaa
...

Collecting factors up to length 8, frequency 7:

Length 2:[2/0] aa, [2/1] bb, [2/2] ab, [2/3] ba
Length 3:[3/0] aaa, [3/1] baa, [3/2] bbb, [3/3] aab, [3/4] abb, [3/5] aba, [3/6] bba
Length 4:[4/0] aaab, [4/1] bbaa, [4/2] abbb, [4/3] abaa, [4/4] aabb, [4/5] bbbb, [4/6] aaaa
Length 5:[5/0] aaabb, [5/1] aabbb, [5/2] bbbaa, [5/3] aabaa, [5/4] aaaba, [5/5] abaab, [5/6] abbbb
Length 6:[6/0] aaabbb, [6/1] aaabaa, [6/2] bbbbaa, [6/3] baabbb, [6/4] aabbbb, [6/5] aaaaab, [6/6] baabaa
Length 7:[7/0] aaabbbb, [7/1] abaabbb, [7/2] aaabaaa, [7/3] aaabaab, [7/4] aaaaabb, [7/5] aabbbaa, [7/6] aaabbba

Considering [length 3 / frequency 1] baa=c.

Step 2

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