Question Number 45840 by Rauny last updated on 17/Oct/18
![a≦7⇒P(!∃x_a )=0, b≦9⇒Q(!∃y_b )=0 for a, b∈N And A⊋A′: A={(x, y)∣P(x)∙Q(y)=0}=A′, B_(∈A) ={(x, y)∈A∣x=y} Then ∀t∈N: ∣B∣=n(t)=f(P(x), Q(y)), also only t can be in [N, M]. find M. :(](Q45840.png)
$${a}\leqq\mathrm{7}\Rightarrow\mathrm{P}\left(!\exists{x}_{{a}} \right)=\mathrm{0}, \\ $$$${b}\leqq\mathrm{9}\Rightarrow\mathrm{Q}\left(!\exists{y}_{{b}} \right)=\mathrm{0}\:\mathrm{for}\:{a},\:{b}\in\mathbb{N} \\ $$$$\mathrm{And}\:{A}\supsetneq{A}':\:{A}=\left\{\left({x},\:{y}\right)\mid\mathrm{P}\left({x}\right)\centerdot\mathrm{Q}\left({y}\right)=\mathrm{0}\right\}={A}', \\ $$$${B}_{\in{A}} =\left\{\left({x},\:{y}\right)\in{A}\mid{x}={y}\right\} \\ $$$$\mathrm{Then}\:\forall{t}\in\mathbb{N}:\:\mid{B}\mid={n}\left({t}\right)={f}\left(\mathrm{P}\left({x}\right),\:\mathrm{Q}\left({y}\right)\right), \\ $$$$\mathrm{also}\:\mathrm{only}\:{t}\:\mathrm{can}\:\mathrm{be}\:\mathrm{in}\:\left[{N},\:{M}\right]. \\ $$$$\mathrm{find}\:{M}. \\ $$$$:\left(\right. \\ $$