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Which of the following propositions is a tautology?

1. $(p \vee q) \rightarrow p$
2. $p \vee (q \rightarrow p)$
3. $p \vee (p \rightarrow q)$
4. $p \rightarrow (p \rightarrow q)$

$A. \ \ (p \vee q) \rightarrow p$
$\quad \quad \equiv \ \ \neg(p \vee q) \vee p$
$\quad \quad \equiv \ \ (\neg p \wedge \neg q) \vee p$
$\quad \quad \equiv \ \ (\neg p \vee p)\wedge(\neg q \vee p)$
$\quad \quad \equiv (p \vee\neg q)$

$B. \ \ p \vee (q \rightarrow p)$
$\quad \quad \equiv \ \ p \vee (\neg q \vee p)$
$\quad \quad \equiv \ \ (p \vee p)\vee\neg q$
$\quad \quad \equiv p \vee\neg q$

$C. \ \ p \vee (p \rightarrow q)$
$\quad \quad \equiv \ \ p \vee (\neg p \vee q)$
$\quad \quad \equiv \ \ ( p \vee \neg p) \vee q$
$\quad \quad \equiv \ \ T \vee q$
$\quad \quad \equiv \ \ T$

$D. \ \ p \rightarrow (p \rightarrow q)$
$\quad \quad \equiv \ \ p \rightarrow (\neg p \vee q)$
$\quad \quad \equiv \ \ \neg p \vee(\neg p \vee q)$
$\quad \quad \equiv \ \ (\neg p \vee\neg p) \vee q$
$\quad \quad \equiv \ \ \neg p \vee q$

Hence, Option(C) $p \vee (p \rightarrow q)$.

C) P OR (P -->Q) = P OR (NOT P OR Q)

=(P OR NOT P) OR Q                  Associativity rule

=T OR Q

=T

### 1 comment

Since only p and q are there, we can also give 1/0 to them and see that all other options can give F.
[c]

p v (p->q) can be written as

p+(p' + q)

p+p'+q

1+q

1

hence  it's a tautology

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