User:KBlott/Proofs/Semilattice with Identity is an Idempotent Commutative Monoid

An algebra $(S, \cdot)$ is a semilattice with identity $e \in S$ iff $(S, \cdot, e)$ is an idempotent commutative monoid.

Proof

$(S, \cdot)$ is a semilattice with identity $e \in S$ iff

• $(S, \cdot)$ is a commutative semigroup, and
• $\forall x \in S$:
  • $x \cdot x = x$, and
  • $x \cdot e = x = e \cdot x$.

Likewise,$(S, \cdot, e)$ is an idempotent commutative monoid iff

• $(S, \cdot)$ is a commutative semigroup, and
• $\forall x \in S$:
  • $x \cdot e = x = e \cdot x$, and
  • $x \cdot x = x$.

$\blacksquare$

Examples and some non-examples

• $(\lbrace 0, \N \rbrace, \cup)$ is a semilattice with identity $0 \in \lbrace 0, \N \rbrace$.
• $(\lbrace 0, \N \rbrace, \cap)$ is a semilattice with identity $\N \in \lbrace 0, \N \rbrace$.
• $(\N, \cup)$ is a semilattice with identity $0 \in \N$.
• $(\N, \cap)$ is a semilattice without identity, since $\N \not \in \N$. This can be fixed by taking the successor set $\N + 1 := \N \cup \lbrace \N \rbrace$.
• $(\N + 1, \cup)$ semilattice with identity $0 \in \N + 1$.
• $(\N + 1, \cap)$ semilattice with identity $\N \in \N + 1$. $\N + 1$ is a subset of the power set of $\N$. This allows us to further generalize $\N$ without losing structural integrity.
• $(\mathscr P(\N), \cup)$ semilattice with identity $0 \in \mathscr P(\N)$ where $\mathscr P(\N)$ is the power set of $\N$.
• $(\mathscr P(\N), \cap)$ semilattice with identity $\N \in \mathscr P(\N)$.
• $(\N, +, 0)$ and $(\N, \times, 1)$ are commutative monoids but not semilattices, since they are not idempotent.
• $(1, +, 0)$ is a semilattice with identity $0 \in 1$.
• $(2, \times, 1)$ is a semilattice with identity $1 \in 2$.

Sources

• Proof Theoretical Studies on Semilattice Relevant Logics. 2001. Ryo Kashim.