#342 ⟨a, b, c | aab=c, ccb=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 ⋅ aba²b²a = a and a ⋅ 1 = a, however aba²b²a ≠ 1
Not right cancellative, because right multiplication by aba²b² is not injective:
- aba²b²a ⋅ aba²b² = aba²b² and 1 ⋅ aba²b² = aba²b², however aba²b²a ≠ 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.	(a²b)²b ⇒ 1	[2]
2.	c ⇒ a²b	[1]

# abc:aab=c,ccb=1 ab/c - -
aabaabb=1
c=aab

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

2 total

Σ	#	Presentation	Mapping
8	4749	⟨a, b, c \| aa=b, bcbcc=1⟩	φ(a) = a, φ(b) = aa, φ(c) = b
8	5037	⟨a, b, c \| ab=c, acacb=1⟩	φ(a) = a, φ(b) = b, φ(c) = ab