Heyting algebra
pseudocomplemented lattice のサブクラス？wint.icon
例を作りたいwint.icon
Nick Bezhanishvili; Dick de Jongh. (2006). "Intuitionistic logic."
Figure 5 (p. 31)
https://gyazo.com/6b89a35f40e1d517ddf4718df1ca99ca
Theorem 47 (p. 31)
A lattice L is distributive ⇔ $ N_5 and $ M_5 are not sublattices of L.
Figure 6 (p. 33)
https://gyazo.com/6af13e5e66d33688e0a5e13e02ec0c6a
Example 52. (p. 33)
1. Every finite distributive lattice is a Heyting algebra. This immediately follows from Proposition 51(2), since every finite distributive lattice is complete and satisfies the infinite distributive law.
Prop 51(2). A complete distributive lattice is a Heyting algebra iff it satisfies the infinite distributive law on V.
2. Every chain C with a least and greatest element is a Heyting algebra and for every a, b ∈ C we have $ a → b = \left\{\begin{array}{} 1 & \rm{if} & a ≤ b, \\ b & \rm{if} & a > b. \end{array}\right.
3. Every Boolean algebra B is a Heyting algebra, where for every a, b ∈ B we have a → b = ¬a ∨ b.
Rieger–Nishimura束
https://upload.wikimedia.org/wikipedia/commons/5/5c/Rieger-Nishimura.svg
明らかに $ p \lor \neg p \neq \top（排中律が成り立たない） 具体例が作れない
by muratak
Bezhanishvili, Nick, and Dick de Jongh. "Intuitionistic logic." (2006).
Example 52
Figure 6
有限 Heyting 代数の例
Theorem 47
Example 52.1
無限 Heyting 代数の例
Rieger–Nishimura lattice (pp. 24–25)