- The binary operation $\cdot$ is associative, that is, $(a\cdot b) \cdot c=a\cdot (b \cdot c)$ for all $a,b,c \in S$.
- There is an identity element $e \in S$, that is, there exists $e \in S$ such that $e \cdot a=a$ and $a \cdot e=a$ for all $a \in S$.
Question: Suppose, $x,y \in S$ such that $x \cdot y=e$. Does $y \cdot x=e$?
This question was motivated by the question here, where the author attempts to prove a special case of the above in the context of matrix multiplication. It was subsequently proved, but the proofs require the properties of the matrix.
I attempted to use Prover9 to prove the statement. Here's the input:
formulas(assumptions). % associativity (x * y) * z = x * (y * z). % identity element a x * a = x. a * x = x. end_of_list. formulas(goals). x * y = a -> y * x = a. end_of_list.
and it returned
sos_empty, which, I guess, implies that no proof of the above statement is possible from the axioms of monoids alone. I ran Mace4 on the same input, and found no counter-examples for monoids of sizes $1,2,\ldots,82$.
A comment by Martin Brandenburg here regarding K-algebras might also apply here. For example, the property might be true for finite monoids, but not all infinite monoids. A counter-example would (obviously) need to be non-commutative.