#171 ⟨a, b, c | aa=1, bbbc=1⟩

Properties

Sum of relation sides is 6
Infinite non-cancellative non-commutative monoid
Element ba has infinite order
Not left cancellative, because left multiplication by b is not injective:
- b ⋅ b²cb = b and b ⋅ 1 = b, however b²cb ≠ 1
Not right cancellative, because right multiplication by b²c is not injective:
- b²cb ⋅ b²c = b²c and 1 ⋅ b²c = b²c, however b²cb ≠ 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.	b³c ⇒ 1	[2]

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

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

7 total

Σ	#	Presentation	Mapping
8	2438	⟨a, b, c \| aab=b, bbbc=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
8	6125	⟨a, b, c \| aa=1, aabbbc=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
8	6162	⟨a, b, c \| aa=1, abbbca=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
8	6198	⟨a, b, c \| aa=1, baabbc=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
8	6234	⟨a, b, c \| aa=1, bbaabc=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
8	6663	⟨a, b, c \| aa=1, abbbc=a⟩	φ(a) = a, φ(b) = b, φ(c) = c
8	7241	⟨a, b, c \| aa=1, bbbc=aa⟩	φ(a) = a, φ(b) = b, φ(c) = c