#855 ⟨a, b, c | ab=c, abcb=1⟩

Properties

Sum of relation sides is 7
Infinite non-cancellative non-commutative monoid
Element a has infinite order
Not left cancellative, because left multiplication by a is not injective:
- a ⋅ bab²a = a and a ⋅ 1 = a, however bab²a ≠ 1
Not right cancellative, because right multiplication by bab² is not injective:
- bab²a ⋅ bab² = bab² and 1 ⋅ bab² = bab², however bab²a ≠ 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

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

#	Rule	Proof
1.	abab² ⇒ 1	[2]
2.	c ⇒ ab	[1]

# abc:ab=c,abcb=1 ab/c - -
ababb=1
c=ab

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

5 total

Σ	#	Presentation	Mapping
7	1376	⟨a, b, c \| ab=1, cacc=b⟩	φ(a) = a, φ(b) = babb, φ(c) = b
8	1845	⟨a, b, c \| aba=ac, cca=1⟩	φ(a) = b, φ(b) = a, φ(c) = ab
8	4700	⟨a, b, c \| aa=b, acacc=1⟩	φ(a) = a, φ(b) = aa, φ(c) = b
8	4920	⟨a, b, c \| ab=a, cacaa=1⟩	φ(a) = b, φ(b) = 1, φ(c) = a
8	5058	⟨a, b, c \| ab=c, babaa=1⟩	φ(a) = b, φ(b) = a, φ(c) = ba