展开 (\forall y)(\exists x)(x=y+1)

解答

展开

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

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

求解步骤

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

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

使用分配律:

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)

化简

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

e x p an d \frac{x ^{2} ( x + 2 z )}{z \cdot y}

e x p an d so l v e (2 x^{2} + 9 x - 5 = 0)

e x p an d \frac{9 x ^{3}}{( x - 6 ) ^{2}}

e x p an d \frac{3 ( x + 1 )}{x [ 2 ]}

e x p an d 5 - (9 x + 5 = 3 - x)