#19 ⟨a, b, c | aa=1, bac=1⟩

Properties

Sum of relation sides is 5
Infinite non-cancellative non-commutative monoid
Element ab has infinite order
Not left cancellative, because left multiplication by b is not injective:
- b ⋅ acb = b and b ⋅ 1 = b, however acb ≠ 1
Not right cancellative, because right multiplication by ac is not injective:
- acb ⋅ ac = ac and 1 ⋅ ac = ac, however acb ≠ 1

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

#	Rule	Proof
1.	a² ⇒ 1	[1]
2.	bac ⇒ 1	[2]

# abc:aa=1,bac=1 abc - -
aa=1
bac=1

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

8 total

Σ	#	Presentation	Mapping
7	305	⟨a, b, c \| aab=b, bac=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1066	⟨a, b, c \| aa=1, aabac=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1073	⟨a, b, c \| aa=1, abaca=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1088	⟨a, b, c \| aa=1, baaac=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1232	⟨a, b, c \| aa=1, abac=a⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1411	⟨a, b, c \| aa=1, bac=aa⟩	φ(a) = a, φ(b) = b, φ(c) = c
8	4025	⟨a, b, c \| abc=1, abcbb=1⟩	φ(a) = b, φ(b) = a, φ(c) = c
8	4059	⟨a, b, c \| abc=1, babcb=1⟩	φ(a) = b, φ(b) = a, φ(c) = c