How to rewrite ~∀x((∀y∀z P(x,y,z))-->(∃z∃y R(x,y,z))) so that the negations appear only within prediates?
That is no negation is outside a quantifier or an expression involving logical connectives. I don't see how you can do that? Is it asking how to get the NOT in front of the P? I tried using distributive law but now I'm just confusing myself.. Any help would be great. If you could list the laws you use that would be great as well I actually want to understand this
I posted this earlier but didn't get a response any ideas would be very much appreciated
I have noticed many times that when you have to puch a negative thru, it tends to flip the operator
-Ax P(x) = Ex -P(x)
\[-\int_{a}^{b}f(x)dx=\int_{b}^{a}f(x)dx\]
\[-(P\ \cap\ Q )=(-P\ \cup\ -Q )\]
I tried that before but I only got this far ∃x ~((∀y∀z P(x,y,z))
then didn't know how to proceed from there I think the question wants me to get the ~ infront of the P
change the innards from p->q to its equivalent -p v q
then a negation across it all will look more doable
Okay I got ∃x~(~(∀y∀z P(x,y,z)) v (∃z∃y R(x,y,z)))
can I multiply in the ~ using distributive law?
∃x((∀y∀z P(x,y,z)) ^ ~(∃z∃y R(x,y,z)))
\[-∀x((∀y∀z P(x,y,z))\rightarrow(∃z∃y R(x,y,z)))\] \[-∀x(-(∀y∀z P(x,y,z))\cup\ (∃z∃y R(x,y,z)))\] \[Ex\ -(-(∀y∀z P(x,y,z))\cup\ (∃z∃y R(x,y,z)))\] \[Ex\ (--(∀y∀z P(x,y,z))-\cup\ -(∃z∃y R(x,y,z)))\] \[Ex\ ((∀y∀z P(x,y,z))\cap\ (-∃z∃y R(x,y,z)))\] \[Ex\ ((∀y∀z P(x,y,z))\cap\ (Az-∃y R(x,y,z)))\] \[Ex\ ((∀y∀z P(x,y,z))\cap\ (AzAy -R(x,y,z)))\] perhaps
Oh snap is that a union sign?
and and ors :)
? the U and then the upside down U
yeah, they come out better in the latex code when trying to do ^ and v
same directions, same meaning, just rounded ends
Ohhh I see so the U is or and the upsidedown U is and?
yep
sorry I started learning about U signs to in class about sets or something so I got confused thanks again!
your welcome
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