1. You have two predicates $p$ and $q$. Express the following situations in first order logic:

    1. there is no \(p\) that is not also \(q\). \(\forall x.px\cond qx\)

    2. there is exactly one \(p\). \(\exists x.px\land \forall y. py \cond y = x\)

    3. there are exactly two $p$s. \(\exists x\exists y. px \land py \land \neg(x = y)\land \forall z. pz \cond z = x\lor z=y\)

    4. there is at most one \(p\). \((\neg\exists x.px) \lor \exists x.px\land \forall y. py \cond y = x\)