Question
Determine the validity of the following arguments using the direct method of truth table:
$M \rightarrow N$
$\sim N$
$\therefore \sim M$

Answer

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

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