Tag: different forms of theoretical statements

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Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Let p and q be any two logical statements and $r : p \rightarrow (\sim p \vee q)$. If r has a truth value F, then the truth values of p and q are respectively

  1. F, F

  2. T, T

  3. F, T

  4. T, F
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
p q $\sim$p $\sim$ p $\vee$ q r
T T F T T
F F T T T
T F F F F
F T T T T

$\therefore$ Clearly from above able, If r has a truth value F, then the truth values of p and core T and F respectively.

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

State, whether the is given the statement, is True or False.
$\sim [(p \vee \sim q) \rightarrow (p \wedge \sim q)] \equiv (p \vee \sim q) \wedge (\sim \vee q)$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 $p$  $q$  $\sim q$  $(p\vee \sim q)$  $(p\wedge \sim q)$ $(p\vee \sim q)\rightarrow (p\wedge \sim q)$  $\sim[(p\vee \sim q)\rightarrow (p\wedge \sim q)]$
 $T$  $T$  $F$  $T$  $F$  $F$  $T$
 $T$  $F$  $T$  $T$  $T$  $T$   $F$
 $F$  $T$  $F$  $F$  $F$  $T$   $F$
 $F$  $F$  $T$  $T$  $F$  $F$ $T$
 $p$  $q$  $\sim q$  $(p\vee \sim q)$ $\sim p$  $(\sim p \vee q)$ $(p\vee \sim q)\wedge (\sim p \vee q)$
 $T$  $T$  $F$  $T$  $F$  $T$  $T$
 $T$  $F$  $T$  $T$  $F$  $F$  $F$
 $F$  $T$  $F$   $F$  $T$  $T$  $F$
 $F$  $F$  $T$  $T$  $T$  $T$  $T$
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

State whether the following statements is True or False?
$p \leftrightarrow q \equiv (p \wedge q) \vee (\sim p \wedge \sim q)$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
given statement 
$p\leftrightarrow q=(p\wedge q)\vee (\sim p\wedge \sim q)$
taking RHS
$(p\wedge q)\vee (\sim p\wedge \sim q)$
$(p\wedge q)\vee (\sim(p\wedge  q))$
$(p\wedge q)\wedge(p\wedge  q)$
$p\leftrightarrow q$
It is true
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Let p,q be statements. Negation of statement $p \leftrightarrow  ~ q$, is

  1. $~ q \rightarrow p$
  2. $ ~ p v q$
  3. $p \leftrightarrow q$
  4. $p \rightarrow q$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
p q ~q $p \leftrightarrow ~q$ $~(p \leftrightarrow ~q)$ $~q \rightarrow p$ $~p v q$ $p \leftrightarrow q$ $p \rightarrow q$
T T F F T T T T T
T F T T F T F F F
F T T T F T T F T
F F F F T F T T T