Some pretty tricky formalisations here because the wording is quite obscure, but there is a right answer somewhere I hope! Any help would be very much appreciated; thank you very much in advanced.
Px: x is a logician; Qx: x is smart; Rx: x is slow
- If all logicians are smart then no logician is slow.
- Some logicians are slow but there are no non-smart logicians.
Sx: x is a Beatles song; Tx: x is a song sung by Ringo; Ux: x is great; a: Octopus’s Garden
- All songs of the Beatles, except those sung by Ringo, are great.
- Octopus’s Garden is a Beatles song and is not great and is not sung by Ringo.
My ideas:
- $\forall x (Px \rightarrow Qx) \rightarrow \forall x (Px \rightarrow \neg Rx)$
- $\exists x(Px \land Rx) \land \forall x (\neg Px \rightarrow \neg Qx)$
- $\forall x ((Sx \land \neg Tx) \rightarrow Ux)$
- $Sa \land \neg Ua \land \neg Ta$
You've nailed $(1),(3), (4).$
First translation matches my work immediately below.
$$\forall x (P(x) \rightarrow Q(x))\rightarrow \lnot \exists x (P(x) \land R(x))\tag 1$$
$$\equiv \forall x (P(x) \rightarrow Q(x))\rightarrow \forall x\lnot(P(x) \land R(x))\tag 2$$
$$\equiv \forall x (P(x) \rightarrow Q(x))\rightarrow \forall x(\lnot P(x) \lor \lnot R(x))\tag{3) DeMorgan's on (2 }$$
$$\equiv \forall x (P(x) \rightarrow Q(x))\rightarrow \forall x( P(x) \rightarrow \lnot R(x))\tag 4$$
Your second translation is not quite right. Consider the following argument: We have
$$\exists x\Big((P(x)\land R(x)\Big) \land \lnot \exists x\Big((P(x) \land \lnot Q(x))\Big)\tag{1}$$
$$\equiv \exists x\Big((P(x)\land R(x))\Big) \land \forall x \lnot\Big(P(x)\land \lnot Q(x)\Big)\tag 2$$
$$\equiv \exists x\Big((P(x)\land R(x))\Big) \land \forall x \Big(\lnot P(x)\lor \lnot\lnot Q(x)\Big)\tag 3$$
$$\equiv \exists x\Big((P(x)\land R(x))\Big) \land \forall x \Big( P(x)\rightarrow Q(x)\Big)\tag 4 $$