Translation Verification Mathematical Logic

Question

May you please tell me if my translation to mathematical logic is correct?

Also if I need more than one variable, may you explain me why? I am lost about when I need one variable and when I need more than one.

Any car with good brakes is safe to drive and safe to ride in. So if a car is new and if all new cars have good brakes, then it is safe to ride in.

Premise:

(For all x) [(Cx and Gx) implies (Dx and Rx)]

Conclusion:

(There exists y) (For all x) ([(Cy and Ny) and ((Cx and Nx) implies Gx)] implies (Ry))

Answer 1

Again, your conclusion isn't quite right. First note that when it says 'if a car ...' it may look like this is about some car, but they really say this about any car. It is like: 'A lion is a dangerous animal' ... if you think about it, they really mean that as a universal, not just about some lion.

So, you get:

$\forall x ((Cx \land Nx) \rightarrow (\forall y ((Cy \land Ny) \rightarrow Gy) \rightarrow Rx))$

If you want to take the quantifier to the front, you can just the following general Prenex Laws, so called because they do exactly that: take the quantifiers to the front, and any FOL statement with all quantifiers at the front is said to be in Prenex Normal Form:

$Q \land \exists x P(x) \Leftrightarrow \exists x (Q \land P(x))$

$Q \land \forall x P(x) \Leftrightarrow \forall x (Q \land P(x))$

$Q \lor \exists x P(x) \Leftrightarrow \exists x (Q \lor P(x))$

$Q \lor \forall x P(x) \Leftrightarrow \forall x (Q \lor P(x))$

$Q \rightarrow \exists x P(x) \Leftrightarrow \exists x (Q \rightarrow P(x))$

$Q \rightarrow \forall x P(x) \Leftrightarrow \forall x (Q \rightarrow P(x))$

$\exists x P(x) \rightarrow Q \Leftrightarrow \forall x (P(x) \rightarrow Q)$

$\forall x P(x)\rightarrow Q \Leftrightarrow \exists x (P(x) \rightarrow Q)$

Note that for the last two, the quantifier changes! This is because:

$\exists x P(x) \rightarrow Q \Leftrightarrow $ (rewrite conditional)

$\neg \exists x P(x) \lor Q \Leftrightarrow $ (quantifier negation)

$\forall x \neg P(x) \lor Q \Leftrightarrow $ (Prenex Law for $\lor$)

$\forall x (\neg P(x) \lor Q) \Leftrightarrow $ (rewrite conditional)

$\forall x (P(x) \rightarrow Q)$

So you get:

$\forall x ((Cx \land Nx) \rightarrow (\forall y ((Cy \land Ny) \rightarrow Gy) \rightarrow Rx)) \Leftrightarrow $

$\forall x ((Cx \land Nx) \rightarrow \exists y(((Cy \land Ny) \rightarrow Gy) \rightarrow Rx)) \Leftrightarrow $

$\forall x \exists y ((Cx \land Nx) \rightarrow (((Cy \land Ny) \rightarrow Gy) \rightarrow Rx))$

... and you can certainly not switch the order of quantifiers of different types.