Formalising Sentences in Monadic Predicate Logic

Question

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

1. If all logicians are smart then no logician is slow.
2. 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

3. All songs of the Beatles, except those sung by Ringo, are great.
4. Octopus’s Garden is a Beatles song and is not great and is not sung by Ringo.

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My ideas:

1. $\forall x (Px \rightarrow Qx) \rightarrow \forall x (Px \rightarrow \neg Rx)$
2. $\exists x(Px \land Rx) \land \forall x (\neg Px \rightarrow \neg Qx)$
3. $\forall x ((Sx \land \neg Tx) \rightarrow Ux)$
4. $Sa \land \neg Ua \land \neg Ta$

Answer 1

You've nailed $(1),(3), (4).$

First translation matches my work immediately below.

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$$\forall x (P(x) \rightarrow Q(x))\rightarrow \lnot \exists x (P(x) \land R(x))\tag 1$$

$$\equiv \forall x (P(x) \rightarrow Q(x))\rightarrow \forall x\lnot(P(x) \land R(x))\tag 2$$

$$\equiv \forall x (P(x) \rightarrow Q(x))\rightarrow \forall x(\lnot P(x) \lor \lnot R(x))\tag{3) DeMorgan's on (2 }$$

$$\equiv \forall x (P(x) \rightarrow Q(x))\rightarrow \forall x( P(x) \rightarrow \lnot R(x))\tag 4$$

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Your second translation is not quite right. Consider the following argument: We have

$$\exists x\Big((P(x)\land R(x)\Big) \land \lnot \exists x\Big((P(x) \land \lnot Q(x))\Big)\tag{1}$$

$$\equiv \exists x\Big((P(x)\land R(x))\Big) \land \forall x \lnot\Big(P(x)\land \lnot Q(x)\Big)\tag 2$$

$$\equiv \exists x\Big((P(x)\land R(x))\Big) \land \forall x \Big(\lnot P(x)\lor \lnot\lnot Q(x)\Big)\tag 3$$

$$\equiv \exists x\Big((P(x)\land R(x))\Big) \land \forall x \Big( P(x)\rightarrow Q(x)\Big)\tag 4 $$