∂q = ∂Q ∂pand ∂q= ∂Q ∂p+ ∂Q ;
∂x ∂p ∂x∂y ∂p ∂y∂y

these give

∂p ∂q −∂p ∂q =∂p ∂Q ;
∂x ∂y∂y ∂x∂x ∂y

by hypothesis ∂p/∂x is not identically zero; therefore if the Jacobian determinant of p and q in regard to x and y is zero identically, so is ∂Q/∂y, or Q does not contain y, so that q is expressible as a function of p only. Conversely, such an expression can be seen at once to make the Jacobian of p and q vanish identically.

Passing now to the case of three variables, suppose that the Jacobian determinant of the three functions F, u, v in regard to x, y, z is identically zero. We prove that if u, v are not themselves functionally connected, F is expressible as a function of u and v. Suppose first that the minors of the elements of ∂F/∂x, ∂F/∂y, ∂F/∂z in the determinant are all identically zero, namely the three determinants such as

∂u ∂v −∂u ∂v ;
∂y ∂z∂z ∂y

then by the case of two variables considered above there exist three functional relations. ψ1(u, v, x) = 0, ψ2(u, v, y) = 0, ψ3(u, v, z) = 0, of which the first, for example, follows from the vanishing of

∂u ∂v −∂u ∂v .
∂y ∂z∂z ∂y

We cannot assume that x is absent from ψ1, or y from ψ2, or z from ψ3; but conversely we cannot simultaneously have x entering in ψ1, and y in ψ2, and z in ψ3, or else by elimination of u and v from the three equations ψ1 = 0, ψ2 = 0, ψ3 = 0, we should find a necessary relation connecting the three independent quantities x, y, z; which is absurd. Thus when the three minors of ∂F/∂x, ∂F/∂y, ∂F/∂z in the Jacobian determinant are all zero, there exists a functional relation connecting u and v only. Suppose no such relation to exist; we can then suppose, for example, that

∂u ∂v −∂u ∂v
∂y ∂z∂z ∂y