HI does anyone know how to convert these two statements into propositional logic,??
1."Any person can fool some of the people all of the time,all of the people some of the time,but not all of the people all of the time"
2."peter has at least two children"
i know these two questions are not propositional logic questions but predicated logic ones.But i'm suppose to convent these questions in to propositional logic.
As the question starts with "Any person", the whole statement regards this 'person' and this can not be translated into propositional logic in any good way (i.e. we can not do the quantification needed). The whole statement will be just one propositional variable, say $A$ .
Same thing here, we cant quantify over the children, and we can not express a relation which state that Peter has at least $2$. Thus it becomes just one propositional variable, say $B$.
Boring answers, but you can't do anything better in propositional-calculus (which may be the thing the exercise want to point out).
Edit:
Why can't we write "$p\wedge q\wedge \neg r$" in 1.?
In this case we would have used the fact that $\forall$ distributes over $\wedge$ i.e. that $\forall xP(x) \wedge\forall xQ(x)$ is equivalent with $\forall x(P(x)\wedge Q(x))$. But the way $1.$ is formulated, is in the format $\forall x (P(x)\wedge Q(x) \wedge \neg R(x))$, thus the $\forall$ quantifier is on the outside and we thus can't translate the sentence in a good way.
It can't actually be done, except in the most schematic way which discards the structure.
In the first case, the three sentences can be represented as follows, using the predicate $Fool(x,y,t)$ to mean "$x$ can fool $y$ at time $t$": $$\begin{align} (\forall x)(\exists y)(\forall t)\,Fool(x,y,t) \tag{1a} \\ (\forall x)(\forall y)(\exists t)\,Fool(x,y,t) \tag{1b} \\ \neg(\exists x)(\forall y)(\forall t)\,Fool(x,y,t) \tag{1c} \end{align}$$
Note that (1c) is equivalent to: $$ (\forall x)(\exists y)\neg(\forall t)\, Fool(x,y,t) \tag{1c'} $$ All of these fine distinctions go away if you represent the sentences in propositional logic. The sentences have no logical connectives, so they must be represented as simple propositional variables $p,q,r$ (or $\neg r$, if you prefer — it really doesn't matter) respectively. Notice that (1a) and (1c) together imply that (if there are any people at all, $x$, then) there are at least 2 people ($y$) :) But that can't be deduced from $p,q,r$, and can't even be expressed in propositional logic.
2) "Peter has at least 2 children". If $C(x,y)$ is a predicate meaning that "$x$ has a child $y$", and if $\mathsf{P}$ is a constant denoting Peter, then in first order logic the sentence can be represented by: $$ (\exists x)(\exists y)\,(C(\mathsf{P},x)\land C(\mathsf{P},y)\land x\ne y) \tag{2} $$ In propositional logic, this has to be represented by a single propositional variable $p$: the sentence has no truth-functional connectives, no propositional structure. From (2), we can deduce that Peter has children (at least one child), and that Peter does not have only one child; but $p$ is a black box from which nothing follows except $p$.
It's a bit of a trick question, and as Ove points out in his answer, perhaps its intent is to show the limitations of propositional logic without variables and quantifiers for individuals.
