What is expand (\forall y)(\exists x)(x=y+1) ?

The solution to expand (\forall y)(\exists x)(x=y+1) is \forall \exists x^2y=y+\forall \exists xy

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

expand

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

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

Solution steps

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

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

Apply the distributive law:

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)

Simplify

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

\frac{x ^{2} ( x + 2 z )}{z \cdot y}

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

\frac{9 x ^{3}}{( x - 6 ) ^{2}}

\frac{3 ( x + 1 )}{x [ 2 ]}

5 - (9 x + 5 = 3 - x)

What is expand (\forall y)(\exists x)(x=y+1) ?
The solution to expand (\forall y)(\exists x)(x=y+1) is \forall \exists x^2y=y+\forall \exists xy