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 To decide an argument is $\textit{valid}$ with $n$ distinct premises as $P_1,P_2,...,P_n$ and conclusion $C$, we need to decide whether $(P_1 \wedge P_2 \wedge...\wedge P_n) \rightarrow C$ is tautology or not.       
        
Which of the following statement(s) is/are correct ?

  1. If we may validly infer $S$ from distinct premises $P_1,P_2,...,P_n$ and $R,$ then we may infer $R \rightarrow S$ from $P_1,P_2,...,P_n.$      
  2. "If $A$ wins, then either $B$ or $C$ will place. If $B$ places, then $A$ will not win. If $D$ places, then $C$ will not. $A$ wins. $\textit{Therefore}$, if $A$ wins, $D$ will not place." is a $\textit{valid}$ argument.     
  3. "If $A$ wins, then either $B$ or $C$ will place. If $B$ places, then $A$ will not win. If $D$ places, then $C$ will not. $A$ wins. $\textit{Therefore}$, if $A$ wins, $D$ will not place." is an $\textit{invalid}$ argument.      
  4. If we may validly infer $S$ from premises $P_1,P_2,...,P_n$ and $R,$ then we $\textit{can't}$ infer $R \rightarrow S$ from $P_1,P_2,...,P_n.$

2 Answers

Best answer
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Answer: A, B.

A)
Given: $(P_1 \land P_2 \land .....\land P_n) \land R \rightarrow S$ is a tautology which is equivalent to
${(P_1 \land P_2 \land .....\land P_n)'} \lor {R'} \lor S$ is a tautology.

Now the second part asks us if $(P_1 \land P_2 \land .....\land P_n) \models (R \rightarrow S) $ is valid.
If this argument is valid then $(P_1 \land P_2 \land .....\land P_n) \rightarrow (R \rightarrow S) $ is a tautology.

The above proposition is equivalent to $\underbrace{{(P_1 \land P_2 \land .....\land P_n)'}}_\text{LHS of first implication} \lor \underbrace{{R'}}_\text{LHS of second implication} \lor S$

We know from the given valid argument that this proprosition is a tautology, hence the second argument is valid.

B)

Let A: A wins, B: B places, C: C places, D: D places

Given argument:
$\begin{array}{ r l } & A \rightarrow {B \lor C} \\ & B \rightarrow {A'} \\ & D \rightarrow {C'} \\ & A \\ \hline \therefore & A \rightarrow {D'} \end{array}$

To check the validity of this argument we will try to make all the premises true assuming the conclusion is false.

$A \rightarrow {D'}$ can only be false when $A$ is true and ${D’}$ is false, ie. $D$ is true.

The 4th premise: $A$  is true.
To make the 3rd premise true $C$ will have to be false since $D$ is true.
To make the 2nd premise true $B$ will have to be false.
Now in the 1st premise $A$ is true while $B$ and $C$ both are false, so $B \lor C$ is false. Therefore this implication is false.
So we have failed to make all the premises true while the conclusion is false.

Thus the argument in question is valid.

Options C & D present the opposite case from option B & A respectively. Therefore, these options are incorrect.
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( P1  $\wedge$ P2  $\wedge$ P3 $\wedge$.... $\wedge$ Pn $\wedge$ R) →  S = ( P1' $\vee$ P2' $\vee$ P3' .... $\vee$ Pn' $\vee R' ) $$\vee$ S 

( P1  $\wedge$ P2  $\wedge$ P3 $\wedge$.... $\wedge$ Pn) → ( R →  S) = ( P1' $\vee$ P2' $\vee$ P3' ....$\vee$ Pn' ) $\vee$ (  R'   $\vee$ S ) 

Both are equal, As given 1) is valid then 2) also valid

so, a) is true d) is false

 

   A→ (B $\vee$ C)

   B→ A’ 

   D→ C’

   A

----------------------------------

$\therefore$ A→ D’

Above argument is valid. On making conclusion false , we can’t make all premises as true.

b)is true c) is false.

Ans: A) and B)

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