36 views
Consider the following arguments

$I)\left \{ \forall x\left [ P\left ( x \right )\rightarrow \left \{ Q\left ( x \right ) \wedge S\left ( x \right )\right \} \right ] ,\forall x\left ( P\left ( x \right )\wedge R\left ( x \right ) \right )\right \}\Rightarrow \forall x\left ( R\left ( x \right ) \rightarrow S\left ( x \right )\right )$

$II) \left \{ \forall x\left \{ P\left ( x \right )\vee Q\left ( x \right ) \right \} ,\forall x\left [ \left \{ \sim P\left ( x \right )\wedge Q\left ( x \right ) \right \} \rightarrow R\left ( x \right )\right ]\right \}\Rightarrow \forall x\left \{ \sim R\left ( x \right ) \rightarrow P\left ( x \right )\right \}$

Which of the following are valid?
| 36 views

+1 vote

$I)\left \{ \forall x\left [ P\left ( x \right )\rightarrow \left \{ Q\left ( x \right ) \wedge S\left ( x \right )\right \} \right ] ,\forall x\left ( P\left ( x \right )\wedge R\left ( x \right ) \right )\right \}\Rightarrow \forall x\left ( R\left ( x \right ) \rightarrow S\left ( x \right )\right )$

$\forall x\left ( P\left ( x \right )\wedge R\left ( x \right ) \right )$ //This means $P$ and $R$ are true for all values of $x$ then only this statement will be true.

$\forall x\left [ \sim P \left ( x \right )\vee \{ \ Q\left ( x \right ) \wedge S\left ( x \right )\ \}\ \right ]$ //Since $P'$ is false so $Q$ and $S$ both have to be true for all values of $x$ in order for this statement to be true.

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$\forall x\left ( \sim R\left ( x \right ) \vee S\left ( x \right )\right )$ // since $R'$ is false and $S$ is true for all values of $x$ so this statement would be true for all values of $x$.

Hence 1. is a valid statement.

$II) \left \{ \forall x\left \{ P\left ( x \right )\vee Q\left ( x \right ) \right \} ,\forall x\left [ \left \{ \sim P\left ( x \right )\wedge Q\left ( x \right ) \right \} \rightarrow R\left ( x \right )\right ]\right \}\Rightarrow \forall x\left \{ \sim R\left ( x \right ) \rightarrow P\left ( x \right )\right \}$

$\forall x\left \{ P\left ( x \right )\vee Q\left ( x \right ) \right \}$

 $P$ $Q$ $Q'$ $P \vee Q$ $P \vee Q'$ 0 0 1 0 1 0 1 0 1 0 1 0 1 1 1 1 1 0 1 1

// since $P \vee Q$ should be true for all values of $x$ so this means the highlighted case should not be there

$\forall x\left [ \left \{ P\left ( x \right )\vee \sim Q\left ( x \right ) \right \} \vee R\left ( x \right )\right ]$

 $R$ $P$ $Q$ $Q'$ $P \vee Q'$ $(P \vee Q') \vee R$ 0 0 0 1 1 1 0 0 1 0 0 0 0 1 0 1 1 1 0 1 1 0 1 1 1 0 0 1 1 1 1 0 1 0 0 1 1 1 0 1 1 1 1 1 1 0 1 1

So after removing the highlighted case the table becomes

 $R$ $P$ $Q$ $Q'$ $P \vee Q'$ $(P \vee Q') \vee R$ 0 0 1 0 0 0 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1 0 0 1 1 1 0 1 1 1 1 1 1 0 1 1

Since $(P \vee Q') \vee R$ should be true for all values of $x$ hence the highlighted case should not be there.

So after removing that case the table becomes

 $R$ $P$ $Q$ $Q'$ $P \vee Q'$ $(P \vee Q') \vee R$ 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1 0 0 1 1 1 0 1 1 1 1 1 1 0 1 1

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$\forall x\left \{ R\left ( x \right ) \vee P\left ( x \right )\right \}$ // In the above table we can see that for all cases $P \vee R$ will be true hence this conclusion is true

Hence 2. is a valid statement.

by (3.7k points)
0

@Satbir

U told

Q and S both have to be true for all values of x in order for this statement to be true.

Then how r u concluding "since Râ€² is false and S is true for all values of x so this statement would be true for all values of x." --------> is valid??

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see the other statement $P \wedge R$ is true for all values of $x$

$\implies R$ is true for all values of $x$

$\implies R'$ should be false for all values of $x$

From 1st statement we are getting R' is  false and from second statement we are getting S to be true for all values of $x$
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How r u getting $R'$ false.

$\sim R$ means $R$ is false. isnot it?
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yes $\sim R \equiv R'$

....i want to say that R=1 i.e. true for all values of x.

so what will be R' ? it will be 0 i.e. false for all values of x
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@Satbir

But by only telling S is true, can we say it is for all x??

Because , forall x,  we need to true Q and S both.

right?

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Yes, $Q$ and $S$ both would be true only when $Q$ is also true for all values of $x$ and $S$ is also true for all values of $x$