#26 ⟨a, b, c | ab=1, acc=1⟩

Properties

Sum of relation sides is 5
Infinite non-cancellative non-commutative monoid
Element a has infinite order
Not left cancellative, because left multiplication by a is not injective:
- a ⋅ ba = a and a ⋅ 1 = a, however ba ≠ 1
Not right cancellative, because right multiplication by b is not injective:
- ba ⋅ b = b and 1 ⋅ b = b, however ba ≠ 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.	ab ⇒ 1	[1]
2.	ac² ⇒ 1	[2]

# abc:ab=1,acc=1 abc - -
ab=1
acc=1

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

8 total

Σ	#	Presentation	Mapping
7	347	⟨a, b, c \| aba=a, acc=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	388	⟨a, b, c \| abc=a, baa=1⟩	φ(a) = c, φ(b) = a, φ(c) = b
7	1124	⟨a, b, c \| ab=1, aabcc=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1138	⟨a, b, c \| ab=1, abacc=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1155	⟨a, b, c \| ab=1, acabc=1⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1324	⟨a, b, c \| ab=1, acca=a⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1344	⟨a, b, c \| ab=1, bacc=b⟩	φ(a) = a, φ(b) = b, φ(c) = c
7	1484	⟨a, b, c \| ab=1, acc=ab⟩	φ(a) = a, φ(b) = b, φ(c) = c