#156 ⟨a, b, c | ab=a, cb=a⟩

Properties

Sum of relation sides is 6
Infinite non-cancellative non-commutative monoid
Element b has infinite order
Not left cancellative, because left multiplication by c is not injective:
- c ⋅ b² = cb and c ⋅ b = cb, however b² ≠ b
Not right cancellative, because right multiplication by b is not injective:
- cb ⋅ b = cb and c ⋅ b = cb, however cb ≠ c

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

Reduction order:
- Left-to-right recursive path with deg(b) = deg(c) = 0, b < c; deg(a) = 1
Certificate: derivations of all rewriting rules from the defining relations.

#	Rule	Proof
1.	cb² ⇒ cb	[4]
2.	a ⇒ cb	[2]

# abc:ab=a,cb=a bc/a - -
cbb=cb
a=cb

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

5 total

Σ	#	Presentation	Mapping
7	525	⟨a, b, c \| ac=ab, bb=c⟩	φ(a) = c, φ(b) = b, φ(c) = bb
7	1057	⟨a, b, c \| ab=a, cb=ab⟩	φ(a) = cb, φ(b) = b, φ(c) = c
8	5712	⟨a, b, c \| aa=b, acc=ac⟩	φ(a) = c, φ(b) = cc, φ(c) = b
8	5803	⟨a, b, c \| ab=a, abb=cb⟩	φ(a) = cb, φ(b) = b, φ(c) = c
8	6059	⟨a, b, c \| ab=c, baa=ba⟩	φ(a) = b, φ(b) = c, φ(c) = bc