# Thread: Laws of logic question

1. ## Laws of logic question

hey guys, i have this question.

$\displaystyle (((p \rightarrow r ) \rightarrow p ) \wedge q ) \wedge (-((p \rightarrow r ) \rightarrow p ))$

my answer after simplifing boths sides i got

$\displaystyle (p \wedge q ) \wedge (-p)$

which then using associative law gives

$\displaystyle q \wedge ( p \wedge -p )$

then inverse law

$\displaystyle q \wedge T$

then Annihilation law

$\displaystyle T$

so this is a tautology. Can anyone confirm this is correct? thanks!

EDIT: Sorry just noticed i posted this thread in the wrong section. Needs to be moved to discrete maths

2. Hello jvignacio
Originally Posted by jvignacio
hey guys, i have this question.

$\displaystyle (((p \rightarrow r ) \rightarrow p ) \wedge q ) \wedge (-((p \rightarrow r ) \rightarrow p ))$

my answer after simplifing boths sides i got

$\displaystyle (p \wedge q ) \wedge (-p)$

which then using associative law gives

$\displaystyle q \wedge ( p \wedge -p )$

then inverse law

$\displaystyle q \wedge T$

then Annihilation law

$\displaystyle T$

so this is a tautology. Can anyone confirm this is correct? thanks!

EDIT: Sorry just noticed i posted this thread in the wrong section. Needs to be moved to discrete maths
I assume that $\displaystyle -p$ means $\displaystyle \neg p$ (or $\displaystyle \sim p$) in which case your answer is not correct. $\displaystyle p \land \neg p = F$. So the result is $\displaystyle F$.

3. Helolo, jvignacio!

i have this question.

$\displaystyle \bigg( \left[(p \to r) \to p \right] \wedge q\bigg) \wedge \bigg(\sim [(p \to r ) \to p)]\bigg)$

Let: .$\displaystyle s \:=\:(p\to r)\to p$

Then we have: .$\displaystyle (s \wedge q)\: \wedge \sim\! s$

. . . . . . . . . .$\displaystyle =\;s \wedge (q \:\wedge \sim\! s)$. . Assoc. Property

. . . . . . . . . .$\displaystyle =\; s \wedge (\sim\!s \wedge q)$ . .Comm. Property

. . . . . . . . . .$\displaystyle =\; (s \:\wedge \sim\!s) \wedge q$. . Assoc. Property

. . . . . . . . . .$\displaystyle =\; F \wedge \:q$

. . . . . . . . . .$\displaystyle =\; F$