Theorem: Conditions for Transformation to be Canonical

Let:
:$\ds J_1 \sqbrk {\sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {p_i}_{1 \mathop \le i \mathop \le n} } = \int_a^b \paren {\sum_{i \mathop = 1}^n p_i y_i'-H} \rd x$
:$\ds J_2 \sqbrk {\sequence {Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n} } = \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i'-H^*} \rd x$
be functionals.
Then $\paren {\sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence{p_i}_{1 \mathop \le i \mathop \le n}, H} \to \paren {\sequence{Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n}, H^*}$ is a canonical transformation if:
:$\ds \sum_{i \mathop = 1}^n p_i y_i' - H = \sum_{i \mathop = 1}^n P_i Y_i' - H^* \pm \dfrac {\d \Phi} {\d x}$
and:
:$\ds p_i = \mp \frac {\partial \Phi} {\d y_i}, \quad P_i = \pm \frac {\partial \Phi} {\d Y_i}, \quad H = H^* \mp \frac {\partial \Phi} {\partial x}$

Proof:
By Conditions for Integral Functionals to have same Euler's Equations, functionals:
:$\ds \int_a^b F_1 \rd x$ 
and:
:$\ds \int_a^b F_2 \rd x = \int_a^b \paren {F_1 \pm \frac {\d \Phi} {\d x} } \rd x$ 
have same Euler's equations.
Express the first one in canonical variables $\paren {x, \sequence{y_i}_{1 \mathop \le i \mathop \le n}, \sequence{p_i}_{1 \mathop \le i \mathop \le n}, H}$ and the second one in $\paren {x, \sequence{Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n}, H^*}$:
:$\ds \int_a^b F_1 \rd x = \int_a^b \paren {\sum_{i \mathop = 1}^n p_i y_i' - H} \rd x$
:$\ds \int_a^b F_2 \rd x = \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i' - H^*} \rd x$
However:
:$\ds \int_a^b \paren {F_2 - F_1} \rd x = \int_a^b \pm \frac {\d \Phi} {\d x} \rd x$
Inserting new expressions for $F_1$ and $F_2$ yields:
:$\ds \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i'-H^* - \sum_{i \mathop = 1}^n p_i y_i' + H} \rd x = \int_a^b \pm \frac {\d \Phi} {\d x}\rd x$
This is satisfied, if integrands are equal.
Transform the coordinates to $\tuple {x, \sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {Y_i}_{1 \mathop \le i \mathop \le n} }$ and write out the full derivative of $\Phi$:
{{DefinitionWanted|Rather than link to a result, we need a definition for "full derivative"}}
:$\ds \sum_{i \mathop = 1}^n P_i Y_i' - H^* - \sum_{i \mathop = 1}^n p_i y_i' + H = \pm \frac {\partial \Phi} {\partial x} \pm \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} y_i' \pm \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial Y_i} Y_i'$
Collect terms multiplied by the same the coordinates together:
:$\ds \sum_{i \mathop = 1}^n Y_i' \paren {\pm \frac {\partial \Phi} {\partial Y_i} - P_i} + \sum_{i \mathop = 1}^n y_i' \paren {\pm \frac {\partial \Phi} {\partial y_i} + p_i} + \paren {\pm \frac {\partial \Phi} {\partial x} - H + H^*} = 0$
This has to hold for arbitrary values of independent coordinates $\tuple {x, \sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {Y_i}_{1 \mathop \le i \mathop \le n} }$.
Hence:
:$p_i = \mp \dfrac {\partial \Phi} {\d y_i}, \quad P_i = \pm \dfrac {\partial \Phi} {\d Y_i}, \quad H^* = H \mp \dfrac {\partial \Phi} {\partial x}$
{{qed}}
{{finish|Eventually cover all four generating functions, most probably in different pages}}