展開する (\forall y)(\exists x)(x=y+1)

解

展開する

(\forall y) (\exists x) (x = y + 1)

\forall \exists x^{2} y = y + \forall \exists x y

解答ステップ

(\forall y) (\exists x) (x = y + 1)

= (\forall y) (\exists x) (= x y + 1)

分配法則を適用する:

a (b + c) = ab + a c

a = (\forall y) (\exists x), b = x = y, c = 1

(\forall y) (\exists x) x = y + (\forall y) (\exists x) \cdot 1

x (\forall y) (\exists x) = y + 1 \cdot (\forall y) (\exists x)

簡素化

x (\forall y) (\exists x) = y + 1 \cdot (\forall y) (\exists x) : \forall \exists x^{2} y = y + \forall \exists x y

\forall \exists x^{2} y = y + \forall \exists x y