Question

using laws of logics how to show that
(∼p∧(∼q∧r))∨(q∧r)⋁(p∧r)↔r

Answer 1

Answer on Question #63324 – Math – Discrete Mathematics

Question

Using laws of logics how to show that

$$(\sim p \wedge (\sim q \wedge r)) \vee (q \wedge r) \vee (p \wedge r) \leftrightarrow r.$$

Solution

$$\begin{array}{l} \left(\sim p \wedge (\sim q \wedge r)\right) \vee (q \wedge r) \vee (p \wedge r) \Leftrightarrow \\ \Leftrightarrow \left(\sim p \wedge (\sim q \wedge r)\right) \vee \left((q \wedge r) \vee (p \wedge r)\right) \Leftrightarrow \\ \Leftrightarrow \left(\sim p \wedge (\sim q \wedge r)\right) \vee \left((r \wedge q) \vee (r \wedge p)\right) \Leftrightarrow \\ \end{array}$$ $$\begin{array}{l} \Leftrightarrow \left(\sim p \wedge (\sim q \wedge r)\right) \vee \left((r \wedge q) \vee (r \wedge p)\right) \Leftrightarrow \\ \Leftrightarrow \left(\sim p \wedge (q \wedge r)\right) \vee \left(r \wedge (q \vee p)\right) \Leftrightarrow \\ \Leftrightarrow \left(\left(\sim p \wedge \sim q\right) \wedge r\right) \vee \left((q \vee p) \wedge r\right) \Leftrightarrow \\ \Leftrightarrow \left(\sim (p \vee q) \wedge r\right) \vee \left((q \vee p) \wedge r\right) \Leftrightarrow \\ \Leftrightarrow \left(\sim (p \vee q) \wedge r\right) \vee ((p \vee q) \wedge r) \Leftrightarrow \\ \Leftrightarrow \left(r \wedge \sim (p \vee q)\right) \vee (r \wedge (p \vee q)) \Leftrightarrow \\ \end{array}$$ $$\begin{array}{l} \Leftrightarrow r \wedge \left(\sim (p \vee q) \vee (p \vee q)\right) \Leftrightarrow \\ \Leftrightarrow r \wedge T \Leftrightarrow \\ \Leftrightarrow r \\ \end{array}$$

Associative Law of disjunction

Commutative Law of conjunction

Distributive Law of conjunction over disjunction

Associative Law of conjunction

Commutative Law of conjunction

DeMorgan's Law

Commutative Law of disjunction

Commutative Law of conjunction

Distributive Law of conjunction over disjunction

Law of excluded middle

Redundance Law

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