# Thread: Laws of logic question

1. ## Laws of logic question

hey guys, i have this question.

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

my answer after simplifing boths sides i got

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

which then using associative law gives

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

then inverse law

$q \wedge T$

then Annihilation law

$T$

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.

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

my answer after simplifing boths sides i got

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

which then using associative law gives

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

then inverse law

$q \wedge T$

then Annihilation law

$T$

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 $-p$ means $\neg p$ (or $\sim p$) in which case your answer is not correct. $p \land \neg p = F$. So the result is $F$.

3. Helolo, jvignacio!

i have this question.

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

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

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

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

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

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

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

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