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- In abstract algebra, a monadic Boolean algebra is an algebraic structure with signature 〈A, ·, +, ', 0, 1, ∃〉 of type 〈2,2,1,0,0,1〉, where 〈A, ·, +, ', 0, 1〉 is a Boolean algebra. The prefixed unary operator ∃ denotes the existential quantifier, which satisfies the identities: ∃0 = 0 ∃x ≥ x ∃(x + y) = ∃x + ∃y ∃x∃y = ∃(x∃y). ∃x is the existential closure of x. Dual to ∃ is the unary operator ∀, the universal quantifier, defined as ∀x := (∃x')'. A monadic Boolean algebra has a dual formulation that takes ∀ as primitive and ∃ as defined, so that ∃x := ' . Hence the dual algebra has signature 〈A, ·, +, ', 0, 1, ∀〉, with 〈A, ·, +, ', 0, 1〉 a Boolean algebra, as before. Moreover, ∀ satisfies the following dualized version of the above identities: ∀1 = 1 ∀x ≤ x ∀(xy) = ∀x∀y ∀x + ∀y = ∀(x + ∀y). ∀x is the universal closure of x.
- Monadyczna algebra Boole'a – algebra Boole'a z dodatkowym działaniem jednoargumentowym <math>\exists</math>, które spełnia pewne warunki naśladujące własności kwantyfikatora egzystencjalnego.
- 在抽象代数中,一元布尔代数是带有如下标识(signature)的代数结构 这里的 <A, ·, +, ', 0, 1> 是布尔代数。 前缀一元算子 ∃ 指示存在量词,它满足恒等式: ∃0 = 0 ∃x ≥ x ∃(x + y) = ∃x + ∃y ∃x∃y = ∃(x∃y). ∃x 是 x 的“存在闭包”。对偶于 ∃ 的是一元算子 ∀,它是全称量词,定义为 ∀x := (∃x')'。 一元布尔代数有对偶公式,取 ∀ 为原始,把 ∃ 定义为 ∃x := ' 。所以对偶的代数有标识 <A, ·, +, ', 0, 1, ∀>,带有 <A, ·, +, ', 0, 1> 是布尔代数。此外,∀ 满足上面恒等式的对偶版本: ∀1 = 1 ∀x ≤ x ∀(xy) = ∀x∀y ∀x + ∀y = ∀(x + ∀y). ∀x 是 x 的“全称闭包”。
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- In abstract algebra, a monadic Boolean algebra is an algebraic structure with signature 〈A, ·, +, ', 0, 1, ∃〉 of type 〈2,2,1,0,0,1〉, where 〈A, ·, +, ', 0, 1〉 is a Boolean algebra. The prefixed unary operator ∃ denotes the existential quantifier, which satisfies the identities: ∃0 = 0 ∃x ≥ x ∃(x + y) = ∃x + ∃y ∃x∃y = ∃(x∃y). ∃x is the existential closure of x.
- Monadyczna algebra Boole'a – algebra Boole'a z dodatkowym działaniem jednoargumentowym <math>\exists</math>, które spełnia pewne warunki naśladujące własności kwantyfikatora egzystencjalnego.
- 在抽象代数中,一元布尔代数是带有如下标识(signature)的代数结构 这里的 <A, ·, +, ', 0, 1> 是布尔代数。 前缀一元算子 ∃ 指示存在量词,它满足恒等式: ∃0 = 0 ∃x ≥ x ∃(x + y) = ∃x + ∃y ∃x∃y = ∃(x∃y).
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