Thread: What should i do next? Logic simplification

[(p ∨ q) =⇒ r] =⇒ (~q ∧ r)

~[~(p ∨ q) v r] v (~q ∧ r) 1.Implication law

[(p ∨ q) /\ ~r] v (~q ∧ r) 2.Double Negation and DeMorgan law

I'm stuck at this point.

Originally Posted by Dukeham

[(p ∨ q) =⇒ r] =⇒ (~q ∧ r)
~[~(p ∨ q) v r] v (~q ∧ r) 1.Implication law
[(p ∨ q) /\ ~r] v (~q ∧ r) 2.Double Negation and DeMorgan law
I'm stuck at this point.

Look at the truth-table
What are you trying to simplify?
From the way it is posted it seems you are proving the implication:
$[(p \vee q) \Rightarrow r] \Rightarrow (\neg q \wedge r)$

I need to simplify the proposition as much as possible so it would be equivalent to the original expression.

Example:
[(p ⇒ q) ⇒ (p v q)] 0. Original expression
[~(~p v q) v (p v q)] 1. Implication law
[(p /\ ~q) v (p v q)] 2. Doble negation and DeMorgan law
[(p /\ ~q) vp) v q 3. Associative law
pvq 4. absorption law

Originally Posted by Dukeham

I need to simplify the proposition as much as possible so it would be equivalent to the original expression.

Example:
[(p ⇒ q) ⇒ (p v q)] 0. Original expression
[~(~p v q) v (p v q)] 1. Implication law
[(p /\ ~q) v (p v q)] 2. Doble negation and DeMorgan law
[(p /\ ~q) vp) v q 3. Associative law
pvq 4. absorption law

This is hardly a simplification:
$[(p \vee q) \Rightarrow r] \Rightarrow (\neg q \wedge r)$
$\begin{gathered} \neg [\neg (p \vee q) \vee r] \vee (\neg q \wedge r) \hfill \\
[(p \vee q) \wedge \neg r] \vee (\neg q \wedge r) \hfill \\
[(p \wedge \neg r) \vee (q \wedge \neg r)] \vee (\neg q \wedge r) \hfill \\
\end{gathered}$

You can compare this truth-table with this
to see equivalence.

so the last thing I can apply is distributive law?
If the truth table indicates that they are equivalent, does it mean that the exercise was performed correctly?

Originally Posted by Dukeham

so the last thing I can apply is distributive law?
If the truth table indicates that they are equivalent, does it mean that the exercise was performed correctly?

$\begin{gathered}
*~ [(p \vee q) \wedge \neg r] \vee (\neg q \wedge r) \hfill \\
**~[(p \wedge \neg r) \vee (q \wedge \neg r)] \vee (\neg q \wedge r) \hfill \\
\end{gathered}$
Going from * to **distribution is used,

Thank you very much. I really appreciate it.