#1200 ⟨a, b | aa=1, babb=bb⟩

Properties

Sum of relation sides is 8
Infinite non-cancellative non-commutative monoid
Element ba has infinite order
Not left cancellative, because left multiplication by b is not injective:
- b ⋅ ab² = b² and b ⋅ b = b², however ab² ≠ b
Not right cancellative, because right multiplication by b is not injective:
- bab ⋅ b = b² and b ⋅ b = b², however bab ≠ b

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.	a² ⇒ 1	[1]
2.	bab² ⇒ b²	[2]

# ab:aa=1,babb=bb ab - -
aa=1
babb=bb

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

6 total

Σ	#	Presentation	Mapping
10	9970	⟨a, b \| aa=1, aababb=bb⟩	φ(a) = a, φ(b) = b
10	10034	⟨a, b \| aa=1, baaabb=bb⟩	φ(a) = a, φ(b) = b
10	10277	⟨a, b \| aa=1, ababb=abb⟩	φ(a) = a, φ(b) = b
10	10515	⟨a, b \| aa=1, babb=aabb⟩	φ(a) = a, φ(b) = b
10	10521	⟨a, b \| aa=1, babb=baab⟩	φ(a) = a, φ(b) = b
10	10525	⟨a, b \| aa=1, bbaa=babb⟩	φ(a) = a, φ(b) = b

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

3 total

Σ	#	Presentation	Mapping
10	9977	⟨a, b \| aa=1, aabbab=bb⟩	φ(a) = a, φ(b) = b
10	10038	⟨a, b \| aa=1, baabab=bb⟩	φ(a) = a, φ(b) = b
10	10285	⟨a, b \| aa=1, abbab=abb⟩	φ(a) = a, φ(b) = b