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How to express these in terms of predicates & quantifiers :

  • Some properties are tautologies
  • The negation of a contradiction is a tautology
  • The dis junction of two contingencies can be a tautology.
  • The conjunction of two tautologies is a tautology.

I could find the answer from the answer key in this sequence as:

  • $\exists xT(x)$
  • $\forall x(C(x)\rightarrow T(\neg x)) $
  • $\exists x\exists y(\neg T(x)\wedge \neg C(x) \wedge \neg T(y) \wedge \neg C(y) \wedge T(x\vee y)) $
  • $\forall x\forall y((T(x) \wedge T(y)) \rightarrow T(x\wedge y))$

From Rosen 5th edition

And not at all able to know how did he arrive at this answer

Can anyone help ? !!

Thanks in advance

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The logic sentences are fairly straightforward translations of the English statements, can you elaborate what you're stuck with? Are you unfamiliar with the notation, or perhaps how to translate them back to English? Or are you fine with this and there's something deeper that you're stuck on? – Luke Mathieson Oct 23 '12 at 6:26
Yeah @LukeMathieson but as we can see the Second statement it says " The negation of a contradiction is a tautology " i.e it should be "$\forall x(C(\neg x)\rightarrow T(x)) $" next the Statement but i din't get the logic behind Rosen's answer ... – thinkinbee Oct 23 '12 at 7:33

The variables stand for properties (or propositions). The predicate $T$ is for tautologies, i.e., $T(x)$ means that the property $x$ is a tautology. $C(x)$ means that $x$ is a contradiction.

Now things should be rather straight forward:
$\exists x T(x)$ means "there is a property $x$ such that $x$ is a tautology".

The second line is "for all propositions $x$ such that $x$ is a contradiction, the negation $\neg x$ is a tautology".

The third line is more interesting. I believe the last $\wedge$ should be $\vee$ for disjunction. Then the line can be explained as follows:

What is a contingency? A property that is neither a tautology nor a contradiction. So this line says "there are $x$ and $y$ such that $x$ and $y$ are contingencies and $x\vee y$ is a tautology".

The last line also has a typo, I think. The comma between $x$ and $y$ should be $\wedge$.

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Yeah i edited that typo apologies for that but shouldn't the second statement be " $\forall x(C(\neg x)\rightarrow T(x)) $" as the statement resembles it ?? – thinkinbee Oct 23 '12 at 7:30
@thinkinbee You should distinguish between two meanings of a 'contradiction' term. The one references a contradictory property, while the other is the predicate which determines whether some arbitrary property is a contradiction. Here 'contradiction' term, as well as a 'tautology' term, are used to reference a property, not a predicate: The negation of a contradiction is a tautology. => The negation of a contradictory property is a tautology. => If property is contradictory then its negation is a tautology. => If property is a contradiction then its negation is a tautology. – olek Oct 23 '12 at 11:43

@Stefan Geschke explains your typos and what the intended answers seem to be.

But there is something quite horribly confusing going on here (in the model answers). Suppose we take the seemingly intended fourth answer $\forall x\forall y((T(x) \wedge T(y)) \rightarrow T(x \wedge y))$.

Then the first '$\wedge$' is the truth-functional connective, which when used to conjoin two sentences (open or closed) produces a sentence. But what about the second use of the '$\wedge$'? '$T$' is a predicate which applies to terms (constants, variables, or expressions built up from them by the use of functions applying to terms). So if '$T(x \land y)$' is to be well-formed, here $\land$ must be expressing not a sentence-connective but a function which applies to two terms to produce another term (i.e. '$\land$' will be here a function expression which combined with two terms denoting propositions produces a term denoting the conjunction of those proposition). Note the fundamental type-distinction between wffs which express propositions and terms which might denote them.

To deploy these two differently typed uses of '$\wedge$' in an unexplained way in what is supposed to be an elementary exercise is Very Bad if not Logically Wicked!

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Sorry being Logically Wicked ;-) it was a typo my serious apologies for the same – thinkinbee Oct 23 '12 at 7:39
No, @thinkinbeee, it isn't you as I was chastising, but rather what's going on in the model answers offered to you :-) – Peter Smith Oct 23 '12 at 8:09
yeah @Peter but ultimately it was me who was at fault cos i did a typo when struggling with the various LateX rules that were necessary to be followed :-) but yes now i have managed to correct the typo i did . could you please help me now – thinkinbee Oct 23 '12 at 9:38

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