Question
$\sim P \rightarrow\ \sim Q$
$\sim Q$
$\therefore \sim P$

Answer

Combining the two bases of this argument as a whole, the argument will be as follows:
$(\sim P \rightarrow \sim Q)\ \&\ \sim Q$
$\therefore \sim P$
Truth Table:
  Support Statement The resulting statement
  $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P$ $Q$ $\sim P$ $\sim Q$ $\sim P \rightarrow \sim Q$ $(\sim P \rightarrow \sim Q)\ \&\ \sim Q$ $\sim P$
$1$ $T$ $T$ $F$ $F$ $T$ $F$ $F$
$2$ $T$ $F$ $F$ $T$ $T$ $T^*$ $F^*$
$3$ $F$ $T$ $T$ $F$ $F$ $F$ $T$
$4$ $F$ $F$ $T$ $T$ $T$ $T$ $T$
  $1(\sim )$ $2(\sim )$ $3, 4(\rightarrow)$ $5, 4 (\&)$ As $3$
Judgment of the validity of the argument: A total of seven columns have been formed in the above truth table. In which the column no. $6th$ base statement and column no. $7$ is the introduction of the result statement. Row out of the total four rows of the truth table. The result statement in $2$ is false $‘F’.$ Hence this argument is disproportionate.

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