#343 ⟨a, b, c | aab=c, ccc=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 ⋅ (aba)³ = a and a ⋅ 1 = a, however (aba)³ ≠ 1
Not right cancellative, because right multiplication by ab(a²b)² is not injective:
- (aba)³ ⋅ ab(a²b)² = ab(a²b)² and 1 ⋅ ab(a²b)² = ab(a²b)², however (aba)³ ≠ 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)³ ⇒ 1	[2]
2.	c ⇒ a²b	[1]

# abc:aab=c,ccc=1 ab/c - -
aabaabaab=1
c=aab

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

4 total

Σ	#	Presentation	Mapping
7	567	⟨a, b, c \| aaa=1, abbc=1⟩	φ(a) = aabaab, φ(b) = a, φ(c) = b
8	1794	⟨a, b, c \| aab=cc, ccc=1⟩	φ(a) = a, φ(b) = b, φ(c) = aabaab
8	3605	⟨a, b, c \| aaa=1, aabbc=1⟩	φ(a) = aab, φ(b) = a, φ(c) = b
8	3616	⟨a, b, c \| aaa=1, abbca=1⟩	φ(a) = aab, φ(b) = a, φ(c) = b