⟨a, b | bab=aaa, bbbb=1⟩

Up:

Monoid enumeration

#4095 ⟨a, b | baa=abb, abab=1⟩

#4118 ⟨a, b | bbb=aaa, abab=1⟩

Properties

Sum of relation sides is 10

Infinite non-cancellative non-commutative monoid

Element a has infinite order

Not left cancellative, because left multiplication by d is not injective:

d ⋅ da² = d and d ⋅ 1 = d, however da² ≠ 1

Not right cancellative, because right multiplication by a is not injective:

da² ⋅ a = a and 1 ⋅ a = a, however da² ≠ 1

Rewriting system

Format:	Pretty Plain
Word:
	Enter a word above to compute its normal form. Tips: Lowercase letters stand for generators. Spaces are ignored. Numbers repeat the previous letter, e.g. `b90`.
Strategy:	Leftmost Rightmost
Result:	1 1

Auxiliary generators:
1. c = bbb
2. d = caacaacaac
Reduction order:
- Left-to-right recursive path with deg(d) = deg(a) = 0, d < a; deg(c) = 1; deg(b) = 2
Certificate: derivations of all rewriting rules from the defining relations.
Morphocompletion: how the auxiliary generators were found.

#	Rule	Proof
1.	ad ⇒ da	[26]
2.	d²a² ⇒ d	[33]
3.	da³ ⇒ a	[25]
4.	dcda² ⇒ dc	[36]
5.	acda² ⇒ ac	[31]
6.	d²c ⇒ cd²	[42]
7.	a²c ⇒ dca⁴	[24]
8.	d²ac ⇒ acd²	[34]
9.	cdac ⇒ d²a	[43]
10.	dc²da² ⇒ dc²	[38]
11.	ac² ⇒ cdca⁷	[30]
12.	acdc ⇒ dc²d²a	[44]
13.	c³d ⇒ daca	[32]
14.	c³a ⇒ daca⁴	[28]
15.	c(dc)² ⇒ acd⁶a	[46]
16.	dc³ ⇒ acda	[37]
17.	c⁴ ⇒ 1	[8]
18.	b ⇒ c³	[9]

# ab:bab=aaa,bbbb=1 da/c/b bbb=c,caacaacaac=d morph:3/0,10/0 ad=da ddaa=d daaa=a dcdaa=dc acdaa=ac ddc=cdd aac=dcaaaa ddac=acdd cdac=dda dccdaa=dcc acc=cdcaaaaaaa acdc=dccdda cccd=daca ccca=dacaaaa cdcdc=acdddddda dccc=acda cccc=1 b=ccc

The mapping is from the listed presentation's alphabet to the current rewriting system's alphabet.

Σ	#	Presentation	Mapping
10	5719	⟨a, b \| aaaa=1, abbba=b⟩	φ(a) = c, φ(b) = a

Presentation

Mapping

5719

⟨a, b | aaaa=1, abbba=b⟩

φ(a) = c, φ(b) = a

#4107 ⟨a, b | bab=aaa, bbbb=1⟩

Contents

Properties

Rewriting system

Isomorphic instances

Up:	Monoid enumeration
Prev:	#4095 ⟨a, b \| baa=abb, abab=1⟩
Next:	#4118 ⟨a, b \| bbb=aaa, abab=1⟩