#235 ⟨a, b, c | ab=1, abc=c⟩

Properties

Sum of relation sides is 6
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]

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

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

12 total

Σ	#	Presentation	Mapping
8	6807	⟨a, b, c \| ab=1, aabbc=c⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	6851	⟨a, b, c \| ab=1, ababc=c⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	6882	⟨a, b, c \| ab=1, abcab=c⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	7306	⟨a, b, c \| ab=1, aabc=ac⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	7350	⟨a, b, c \| ab=1, abac=ac⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	7367	⟨a, b, c \| ab=1, abbc=bc⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	7377	⟨a, b, c \| ab=1, abca=ca⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	7387	⟨a, b, c \| ab=1, abcb=cb⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	7397	⟨a, b, c \| ab=1, abcc=cc⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	7480	⟨a, b, c \| ab=1, babc=bc⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	7588	⟨a, b, c \| ab=1, cabc=cc⟩	φ(a) = a, φ(b) = b, φ(c) = 1
8	7821	⟨a, b, c \| ab=1, cab=abc⟩	φ(a) = a, φ(b) = b, φ(c) = 1