量化子(全称命題・存在命題)の命題関数と命題変数への分配
量化子(全称命題・存在命題)の命題関数と命題変数への分配
論理和
論理積
論理包含
否定論理和
否定論理積
否定論理包含
論理和
(1)
\[ \exists x\in X,P\left(x\right)\lor Q\Leftrightarrow\left(\left(\exists x\in X,P\left(x\right)\right)\lor Q\right)\land\left(X\ne\emptyset\right) \](2)
\[ \forall x\in X,P\left(x\right)\lor Q\Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\lor Q \]論理積
(3)
\[ \exists x\in X,P\left(x\right)\land Q\Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\land Q \](4)
\[ \forall x\in X,P\left(x\right)\land Q\Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\land Q\right)\lor\left(X=\emptyset\right) \]論理包含
(5)
\[ \exists x\in X,P\left(x\right)\rightarrow Q\Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\rightarrow Q\right)\land\left(X\ne\emptyset\right) \](6)
\[ \forall x\in X,P\left(x\right)\rightarrow Q\Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\rightarrow Q \](7)
\[ \exists x\in X,P\rightarrow Q\left(x\right)\Leftrightarrow\left(P\rightarrow\left(\exists x\in X,Q\left(x\right)\right)\right)\land\left(X\ne\emptyset\right) \](8)
\[ \forall x\in X,P\rightarrow Q\left(x\right)\Leftrightarrow P\rightarrow\left(\forall x\in X,Q\left(x\right)\right) \]否定論理和
(9)
\[ \exists x\in X,P\left(x\right)\downarrow Q\Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\downarrow Q \](10)
\[ \forall x\in X,P\left(x\right)\downarrow Q\Leftrightarrow\left(\left(\exists x\in X,P\left(x\right)\right)\downarrow Q\right)\lor\left(X=\emptyset\right) \]否定論理積
(11)
\[ \exists x\in X,P\left(x\right)\uparrow Q\Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\uparrow Q\right)\land\left(X\ne\emptyset\right) \](12)
\[ \forall x\in X,P\left(x\right)\uparrow Q\Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\uparrow Q \]否定論理包含
(13)
\[ \exists x\in X,P\left(x\right)\nrightarrow Q\Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\nrightarrow Q \](14)
\[ \forall x\in X,P\left(x\right)\nrightarrow Q\Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\nrightarrow Q\right)\lor\left(X=\emptyset\right) \](15)
\[ \exists x\in X,P\nrightarrow Q\left(x\right)\Leftrightarrow P\nrightarrow\left(\forall x\in X,Q\left(x\right)\right) \](16)
\[ \forall x\in X,P\nrightarrow Q\left(x\right)\Leftrightarrow\left(P\nrightarrow\left(\exists x\in X,Q\left(x\right)\right)\right)\lor\left(X=\emptyset\right) \](1)
論理和と論理積のみ求めればあとはそれを使って求めることができます。また、論理和を求めれば、対偶をとることにより論理積も求めることができます。
(2)
\(P,Q\)ともに命題変数ではなく命題関数\(P\left(x\right),Q\left(x\right)\)のときは\[ \exists x\in X,\left(P\left(x\right)\lor Q\left(x\right)\right)\Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\lor\left(\exists x\in X,Q\left(x\right)\right) \] \[ \forall x\in X,P\left(x\right)\lor Q\left(x\right)\Leftarrow\left(\forall x\in X,P\left(x\right)\right)\lor\left(\forall x\in X,Q\left(x\right)\right) \] \[ \exists x\in X,P\left(x\right)\land Q\left(x\right)\Rightarrow\left(\exists x\in X,P\left(x\right)\right)\land\left(\exists x\in X,Q\left(x\right)\right) \] \[ \forall x\in X,P\left(x\right)\land Q\left(x\right)\Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\land\left(\forall x\in X,Q\left(x\right)\right) \] のようになります。
(1)
\(\Rightarrow\)
\begin{align*} \exists x\in X,P\left(x\right)\lor Q & \Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\lor\left(\exists x\in X,Q\right)\cmt{\because\exists x\left(P\left(x\right)\lor Q\left(x\right)\right)\Leftrightarrow\exists xP\left(x\right)\lor\exists xQ\left(x\right)}\\ & \Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\lor\left(Q\land\left(X\ne\emptyset\right)\right)\cmt{\because\exists x\in X,Q\Leftrightarrow Q\land\left(X\ne\emptyset\right)}\\ & \Leftrightarrow\left(\exists x\in X,P\left(x\right)\land\left(X\ne\emptyset\right)\right)\lor\left(Q\land\left(X\ne\emptyset\right)\right)\cmt{\because\exists x\in X,P\left(x\right)\Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\land\left(X\ne\emptyset\right)}\\ & \Leftrightarrow\left(\left(\exists x\in X,P\left(x\right)\right)\lor Q\right)\land\left(X\ne\emptyset\right) \end{align*}(2)
\begin{align*} \forall x\in X,P\left(x\right)\lor Q & \Leftrightarrow\begin{cases} \forall x\in\emptyset,P\left(x\right)\lor Q & X=\emptyset\\ \forall x\in X,P\left(x\right)\lor Q & X\ne\emptyset \end{cases}\\ & \Leftrightarrow\begin{cases} \top & X=\emptyset\\ \left(\forall x\in X,P\left(x\right)\right)\lor Q & X\ne\emptyset \end{cases}\\ & \Leftrightarrow\begin{cases} \top\lor Q & X=\emptyset\\ \left(\forall x\in X,P\left(x\right)\right)\lor Q & X\ne\emptyset \end{cases}\\ & \Leftrightarrow\begin{cases} \left(\forall x\in\emptyset,P\left(x\right)\right)\lor Q & X=\emptyset\\ \left(\forall x\in X,P\left(x\right)\right)\lor Q & X\ne\emptyset \end{cases}\\ & \Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\lor Q \end{align*} となる。従って、題意は成り立つ。
(3)
(2)より、\begin{align*} \exists x\in X,P\left(x\right)\land Q & \Leftrightarrow\lnot\left(\forall x\in X,\lnot P\left(x\right)\lor\lnot Q\right)\\ & \Leftrightarrow\lnot\left(\left(\forall x\in X,\lnot P\left(x\right)\right)\lor\lnot Q\right)\cmt{\because\forall x\in X,P\left(x\right)\lor Q\Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\lor Q}\\ & \Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\land Q \end{align*}
(4)
\begin{align*} \forall x\in X,P\left(x\right)\land Q & \Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\land\left(\forall x\in X,Q\right)\cmt{\because\forall x\left(P\left(x\right)\land Q\left(x\right)\right)\Leftrightarrow\forall xP\left(x\right)\land\forall xQ\left(x\right)}\\ & \Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\land\left(Q\lor\left(X=\emptyset\right)\right)\cmt{\because\forall x\in X,Q\Leftrightarrow Q\lor\left(X=\emptyset\right)}\\ & \Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\lor\left(X=\emptyset\right)\right)\land\left(Q\lor\left(X=\emptyset\right)\right)\cmt{\because\forall x\in X,P\left(x\right)\Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\lor\left(X=\emptyset\right)}\\ & \Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\land Q\right)\lor\left(X=\emptyset\right) \end{align*}(5)
(1)より、\begin{align*} \exists x\in X,P\left(x\right)\rightarrow Q & \Leftrightarrow\exists x\in X,\lnot P\left(x\right)\lor Q\\ & \Leftrightarrow\left(\left(\exists x\in X,\lnot P\left(x\right)\right)\lor Q\right)\land\left(X\ne\emptyset\right)\cmt{\because\exists x\in X,P\left(x\right)\lor Q\Leftrightarrow\left(\left(\exists x\in X,P\left(x\right)\right)\lor Q\right)\land\left(X\ne\emptyset\right)}\\ & \Leftrightarrow\left(\lnot\left(\forall x\in X,P\left(x\right)\right)\lor Q\right)\land\left(X\ne\emptyset\right)\\ & \Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\rightarrow Q\right)\land\left(X\ne\emptyset\right) \end{align*}
(6)
(2)より、\begin{align*} \forall x\in X,P\left(x\right)\rightarrow Q & \Leftrightarrow\forall x\in X,\lnot P\left(x\right)\lor Q\\ & \Leftrightarrow\left(\forall x\in X,\lnot P\left(x\right)\right)\lor Q\cmt{\because\forall x\in X,P\left(x\right)\lor Q\Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\lor Q}\\ & \Leftrightarrow\lnot\left(\exists x\in X,P\left(x\right)\right)\lor Q\\ & \Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\rightarrow Q \end{align*}
(7)
(1)より、\begin{align*} \exists x\in X,P\rightarrow Q\left(x\right) & \Leftrightarrow\exists x\in X,\lnot P\lor Q\left(x\right)\\ & \Leftrightarrow\left(\left(\exists x\in X,Q\left(x\right)\right)\lor\lnot P\right)\land\left(X\ne\emptyset\right)\cmt{\because\exists x\in X,P\left(x\right)\lor Q\Leftrightarrow\left(\left(\exists x\in X,P\left(x\right)\right)\lor Q\right)\land\left(X\ne\emptyset\right)}\\ & \Leftrightarrow\left(\left(\exists x\in X,Q\left(x\right)\right)\leftarrow P\right)\land\left(X\ne\emptyset\right)\\ & \Leftrightarrow\left(P\rightarrow\left(\exists x\in X,Q\left(x\right)\right)\right)\land\left(X\ne\emptyset\right) \end{align*}
(8)
(2)より、\begin{align*} \forall x\in X,P\rightarrow Q\left(x\right) & \Leftrightarrow\forall x\in X,\lnot P\lor Q\left(x\right)\\ & \Leftrightarrow\left(\forall x\in X,Q\left(x\right)\right)\lor\lnot P\cmt{\because\forall x\in X,P\left(x\right)\lor Q\Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\lor Q}\\ & \Leftrightarrow\left(\forall x\in X,Q\left(x\right)\right)\leftarrow P\\ & \Leftrightarrow P\rightarrow\left(\forall x\in X,Q\left(x\right)\right) \end{align*}
(9)
(3)より、\begin{align*} \exists x\in X,P\left(x\right)\downarrow Q & \Leftrightarrow\exists x\in X,\lnot P\left(x\right)\land\lnot Q\\ & \Leftrightarrow\left(\exists x\in X,\lnot P\left(x\right)\right)\land\lnot Q\cmt{\because\exists x\in X,P\left(x\right)\land Q\Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\land Q}\\ & \Leftrightarrow\lnot\left(\forall x\in X,P\left(x\right)\right)\land\lnot Q\\ & \Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\downarrow Q \end{align*}
(10)
(4)より、\begin{align*} \forall x\in X,P\left(x\right)\downarrow Q & \Leftrightarrow\forall x\in X,\lnot P\left(x\right)\land\lnot Q\\ & \Leftrightarrow\left(\left(\forall x\in X,\lnot P\left(x\right)\right)\land\lnot Q\right)\lor\left(X=\emptyset\right)\cmt{\because\forall x\in X,P\left(x\right)\land Q\Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\land Q\right)\lor\left(X=\emptyset\right)}\\ & \Leftrightarrow\left(\lnot\left(\exists x\in X,P\left(x\right)\right)\land\lnot Q\right)\lor\left(X=\emptyset\right)\\ & \Leftrightarrow\left(\left(\exists x\in X,P\left(x\right)\right)\downarrow Q\right)\lor\left(X=\emptyset\right) \end{align*}
(11)
(1)より、\begin{align*} \exists x\in X,P\left(x\right)\uparrow Q & \Leftrightarrow\exists x\in X,\lnot P\left(x\right)\lor\lnot Q\\ & \Leftrightarrow\left(\left(\exists x\in X,\lnot P\left(x\right)\right)\lor\lnot Q\right)\land\left(X\ne\emptyset\right)\cmt{\because\exists x\in X,P\left(x\right)\lor Q\Leftrightarrow\left(\left(\exists x\in X,P\left(x\right)\right)\lor Q\right)\land\left(X\ne\emptyset\right)}\\ & \Leftrightarrow\left(\lnot\left(\forall x\in X,P\left(x\right)\right)\lor\lnot Q\right)\land\left(X\ne\emptyset\right)\\ & \Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\uparrow Q\right)\land\left(X\ne\emptyset\right) \end{align*}
(12)
(2)より、\begin{align*} \forall x\in X,P\left(x\right)\uparrow Q & \Leftrightarrow\forall x\in X,\lnot P\left(x\right)\lor\lnot Q\\ & \Leftrightarrow\left(\forall x\in X,\lnot P\left(x\right)\right)\lor\lnot Q\cmt{\because\forall x\in X,P\left(x\right)\lor Q\Leftrightarrow\left(\forall x\in X,P\left(x\right)\right)\lor Q}\\ & \Leftrightarrow\lnot\left(\exists x\in X,P\left(x\right)\right)\lor\lnot Q\\ & \Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\uparrow Q \end{align*}
(13)
(3)より、\begin{align*} \exists x\in X,P\left(x\right)\nrightarrow Q & \Leftrightarrow\exists x\in X,P\left(x\right)\land\lnot Q\\ & \Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\land\lnot Q\cmt{\because\exists x\in X,P\left(x\right)\land Q\Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\land Q}\\ & \Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\nrightarrow Q \end{align*}
(14)
(4)より、\begin{align*} \forall x\in X,P\left(x\right)\nrightarrow Q & \Leftrightarrow\forall x\in X,P\left(x\right)\land\lnot Q\\ & \Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\land\lnot Q\right)\lor\left(X=\emptyset\right)\cmt{\because\forall x\in X,P\left(x\right)\land Q\Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\land Q\right)\lor\left(X=\emptyset\right)}\\ & \Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\nrightarrow Q\right)\lor\left(X=\emptyset\right) \end{align*}
(15)
(3)より、\begin{align*} \exists x\in X,P\nrightarrow Q\left(x\right) & \Leftrightarrow\exists x\in X,P\land\lnot Q\left(x\right)\\ & \Leftrightarrow\left(\exists x\in X,\lnot Q\left(x\right)\right)\land P\cmt{\because\exists x\in X,P\left(x\right)\land Q\Leftrightarrow\left(\exists x\in X,P\left(x\right)\right)\land Q}\\ & \Leftrightarrow\lnot\left(\forall x\in X,Q\left(x\right)\right)\land P\\ & \Leftrightarrow\left(\forall x\in X,Q\left(x\right)\right)\nleftarrow P\\ & \Leftrightarrow P\nrightarrow\left(\forall x\in X,Q\left(x\right)\right) \end{align*}
(16)
(4)より、\begin{align*} \forall x\in X,P\nrightarrow Q\left(x\right) & \Leftrightarrow\forall x\in X,P\land\lnot Q\left(x\right)\\ & \Leftrightarrow\left(\left(\forall x\in X,\lnot Q\left(x\right)\right)\land P\right)\lor\left(X=\emptyset\right)\cmt{\because\forall x\in X,P\left(x\right)\land Q\Leftrightarrow\left(\left(\forall x\in X,P\left(x\right)\right)\land Q\right)\lor\left(X=\emptyset\right)}\\ & \Leftrightarrow\left(\lnot\left(\exists x\in X,Q\left(x\right)\right)\land P\right)\lor\left(X=\emptyset\right)\\ & \Leftrightarrow\left(\left(\exists x\in X,Q\left(x\right)\right)\nleftarrow P\right)\lor\left(X=\emptyset\right)\\ & \Leftrightarrow\left(P\nrightarrow\left(\exists x\in X,Q\left(x\right)\right)\right)\lor\left(X=\emptyset\right) \end{align*}
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転換法
量化子(全称命題・存在命題)の順序変更
\[
\exists x\forall y,P\left(x,y\right)\Rightarrow\forall y\exists x,P\left(x,y\right)
\]
量化子(全称命題・存在命題)と和集合
\[
\forall x\in A\cup B,P\left(x\right)\Leftrightarrow\left(\forall x\in A,P\left(x\right)\right)\land\left(\forall x\in B,P\left(x\right)\right)
\]
量化子(全称命題・存在命題)と空集合と命題変数
\[
\forall x\in\emptyset,P\left(x\right)\Leftrightarrow\top
\]