The negation of the statement -prightarrow q-wedge r ispwedge sim qvee sim r-sim pwedge q-wedge -sim r-pwedge sim q-wedge -r-pwedge sim q-wedge -sim r-

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The negation of the statement $(p \to q) \land r$ is
$p \land \sim q \lor \sim r$
$(\sim p \land q) \land (\sim r)$
$(p \land \sim q) \land (r)$
$(p \land \sim q) \land (\sim r)$

$p \to q$
$=\sim p \lor q - - - (1)$
$\sim [(p \to q) \land r]$
$\sim [\sim p \lor q \land r]$ ...[form(i)]
$\sim (\sim p \lor q) \lor \sim r$ ...[Demorgans low]
$\sim (\sim p) \land \sim q \lor q \sim r$ $a v (b \land c)$
$p \land \sim q \lor \sim r$ $= (a \lor b) \lor (a \lor c)]$

The negation of the statement (p→q)∧r isp∧∼q∨∼r(∼p∧q)∧(∼r)(p∧∼q)∧(r)(p∧∼q)∧(∼r)

p→q=∼p∨q−−−(1)∼[(p→q)∧r]∼[∼p∨q∧r]...[form(i)]∼(∼p∨q)∨∼r...[Demorgans low]∼(∼p)∧∼q∨q∼r av(b∧c)p∧∼q∨∼r =(a∨b)∨(a∨c)]

The negation of the statement $(p \to q) \land r$ is
$p \land \sim q \lor \sim r$
$(\sim p \land q) \land (\sim r)$
$(p \land \sim q) \land (r)$
$(p \land \sim q) \land (\sim r)$

$p \to q$
$=\sim p \lor q - - - (1)$
$\sim [(p \to q) \land r]$
$\sim [\sim p \lor q \land r]$ ...[form(i)]
$\sim (\sim p \lor q) \lor \sim r$ ...[Demorgans low]
$\sim (\sim p) \land \sim q \lor q \sim r$ $a v (b \land c)$
$p \land \sim q \lor \sim r$ $= (a \lor b) \lor (a \lor c)]$