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Let
$X,Y$ sets,
$X_1\subseteq X$,
$Y_1\subseteq Y$,
$F=\mathcal{P}(X\times Y)$ the set of all functions in $X\times Y$,
$F_1=\{f\in F\ |\ \forall x_0((x_0\in\text{dom}(f))\land (x_0\in X_1))\colon f(x_0)\in Y_1\}$.

$(\exists f\in F)\land(\exists x_0(x_0\in\text{dom}(f))\land(f(x_0)\notin Y_1))\implies f\notin F_1$.

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Let
$X,Y$ sets,
$X_1\subseteq X$,
$Y_1\subseteq Y$,
$F=\mathcal{P}(X\times Y)$ the set of all functions in $X\times Y$,
$F_1=\{f\in F\ |\ \forall x_0((x_0\in\text{dom}(f))\land (x_0\in X_1))\colon f(x_0)\in Y_1\}$.

$(\exists f\in F)\land(\exists x_0(x_0\in\text{dom}(f))\land(f(x_0\notin Y_1))\implies f\notin F_1$.

Let
$X,Y$ sets,
$X_1\subseteq X$,
$Y_1\subseteq Y$,
$F=\mathcal{P}(X\times Y)$ the set of all functions in $X\times Y$,
$F_1=\{f\in F\ |\ \forall x_0((x_0\in\text{dom}(f))\land (x_0\in X_1))\colon f(x_0)\in Y_1\}$.

$(\exists f\in F)\land(\exists x_0(x_0\in\text{dom}(f))\land(f(x_0)\notin Y_1))\implies f\notin F_1$.

Let
$X,Y$ sets,
$X_1\subseteq X$,
$Y_1\subseteq Y$,
$F=\mathcal{P}(X\times Y)$ the set of all functions in $X\times Y$,
$F_1=\{\tilde{f}\in F\ |\ \forall x_0((x_0\in\text{dom}(\tilde{f}))\land (x_0\in X_1))\colon \tilde{f}(x_0)\in Y_1\}$.

If there is a function $f\in F$ and an $x_0\in\text{dom}(f)$ with $f(x_0)\notin Y_1$ then $f\notin F_1$.

$(\exists f\in F)\land(\exists x_0(x_0\in\text{dom}(f))\land(f(x_0\notin Y_1))\implies f\notin F_1$.

Let
$X,Y$ sets,
$X_1\subseteq X$,
$Y_1\subseteq Y$,
$F=\mathcal{P}(X\times Y)$ the set of all functions in $X\times Y$,
$F_1=\{f\in F\ |\ \forall x_0((x_0\in\text{dom}(f))\land (x_0\in X_1))\colon f(x_0)\in Y_1\}$.

$(\exists f\in F)\land(\exists x_0(x_0\in\text{dom}(f))\land(f(x_0\notin Y_1))\implies f\notin F_1$.