( p V q ) V r
MathBot Answer:
Logic Expression
Symbol: $p \vee q \vee r$
Text: $p \text{ or } q \text{ or } r$
Classification contingency
Truth Table
Legend
$p$ | $q$ | $r$ | $p \vee q \vee r$ |
---|---|---|---|
1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 |
1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 |
0 | 1 | 1 | 1 |
0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 |
0 | 0 | 0 | 0 |
Simplification$$p \vee q \vee r$$
Conjunctive Normal Form$$p \vee q \vee r$$
Note 1: These equivalences and tautologies are used to generate the above steps.
Note 2: Two logical statements are logically equivalent if they always produce the same truth value. Consequently, p ≡ q is same as saying p ↔ q is a tautology. $$\begin{array}{c|c}\textbf{Equivalence} \\ \hline \text{Absorption Law} & \begin{gathered} p \wedge \left(p \vee q\right) \equiv p \\ p \vee \left(p \wedge q\right) \equiv p \end{gathered} \\ \hline \text{Biconditional Equivalence} & \begin{gathered} p \leftrightarrow q \equiv \left(p \vee \neg q\right) \wedge \left(\neg p \vee q\right) \\ p \leftrightarrow q \equiv \left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right) \end{gathered} \\ \hline \text{Biconditional Simplification} & \begin{gathered} p \leftrightarrow p \equiv \text{True} & p \leftrightarrow \text{True} \equiv p \\ p \leftrightarrow \neg p \equiv \text{False} & p \leftrightarrow \text{False} \equiv \neg p \end{gathered} \\ \hline \text{Complement Law} & \begin{gathered} p \wedge \neg p \equiv \text{False} \\ p \vee \neg p \equiv \text{True} \end{gathered} \\ \hline \text{Conditional Equivalence} & p \rightarrow q \equiv \neg p \vee q \\ \hline \text{Conditional Simplification} & \begin{gathered} p \rightarrow p \equiv \text{True} & p \rightarrow \text{True} \equiv \text{True} & p \rightarrow \text{False} \equiv \neg p \\ p \rightarrow \neg p \equiv \neg p & \text{True} \rightarrow p \equiv p & \text{False} \rightarrow p \equiv \text{True} \\ \neg p \rightarrow p \equiv p\end{gathered} \\ \hline \text{Consensus Law} & \begin{gathered} \left(p \vee q\right) \wedge \left(\neg p \vee r\right) \wedge \left(q \vee r\right) \equiv \left(p \vee q\right) \wedge \left(\neg p \vee r\right) \\ \left(p \wedge q\right) \vee \left(\neg p \wedge r\right) \vee \left(q \wedge r\right) \equiv \left(p \wedge q\right) \vee \left(\neg p \wedge r\right) \end{gathered} \\ \hline \text{De Morgan's Law} & \begin{gathered} \neg \left(p \wedge q\right) \equiv \neg p \vee \neg q \\ \neg \left(p \vee q\right) \equiv \neg p \wedge \neg q \end{gathered} \\ \hline \text{Distributive Law} & \begin{gathered} p \wedge \left(q \vee r\right) \equiv \left(p \wedge q\right) \vee \left(p \wedge r\right) \\ p \vee \left(q \wedge r\right) \equiv \left(p \vee q\right) \wedge \left(p \vee r\right) \\ \left(p \vee q\right) \wedge \left(r \vee s\right) \equiv \left(p \wedge r\right) \vee \left(p \wedge s\right) \vee \left(q \wedge r\right) \vee \left(q \wedge s\right) \\ \left(p \wedge q\right) \vee \left(r \wedge s\right) \equiv \left(p \vee r\right) \wedge \left(p \vee s\right) \wedge \left(q \vee r\right) \wedge \left(q \vee s\right) \end{gathered} \\ \hline \text{Domination Law} & \begin{gathered} p \vee \text{True} \equiv \text{True} \\ p \wedge \text{False} \equiv \text{False} \end{gathered} \\ \hline \text{Double Negation Law} & \neg \left(\neg p\right) \equiv p \\ \hline \text{Idempotent Law} & \begin{gathered} p \wedge p \equiv p \\ p \vee p \equiv p \end{gathered} \\ \hline \text{Identity Law} & \begin{gathered} p \wedge \text{True} \equiv p \\ p \vee \text{False} \equiv p \end{gathered} \\ \hline \text{NAND} & p \uparrow q \equiv \neg \left(p \wedge q\right) \\ \hline \text{Negation Law} & \begin{gathered} \neg \text{True} \equiv \text{False} \\ \neg \text{False} \equiv \text{True} \end{gathered} \\ \hline \text{NOR} & p \downarrow q \equiv \neg \left(p \vee q\right) \\ \hline \text{Negation of Biconditional Equivalence} & \begin{gathered} \neg \left(p \leftrightarrow q\right) \equiv \left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right) \\ \neg \left(p \leftrightarrow q\right) \equiv \left(p \wedge \neg q\right) \vee \left(\neg p \wedge q\right) \end{gathered} \\ \hline \text{Negation of Conditional Equivalence} & \neg \left(p \rightarrow q\right) \equiv p \wedge \neg q \\ \hline \text{Redundancy Law (1)} & \begin{gathered} \left(p \vee q\right) \wedge \left(p \vee \neg q\right) \equiv p \\ \left(p \wedge q\right) \vee \left(p \wedge \neg q\right) \equiv p \end{gathered} \\ \hline \text{Redundancy Law (2)} & \begin{gathered} p \wedge \left(\neg p \vee q\right) \equiv p \wedge q \\ p \vee \left(\neg p \wedge q\right) \equiv p \vee q \end{gathered} \\ \hline \text{XOR} & \begin{gathered} p \oplus q \equiv \left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right) \\ p \oplus q \equiv \left(p \wedge \neg q\right) \vee \left(\neg p \wedge q\right) \end{gathered} \\ \hline \text{XOR Simplification} & \begin{gathered} p \oplus p \equiv \text{False} & p \oplus \text{True} \equiv \neg p \\ p \oplus \neg p \equiv \text{True} & p \oplus \text{False} \equiv p \end{gathered} \\ \hline \text{XNOR} & p \odot q \equiv \neg \left(p \oplus q\right) \end{array}$$ $$\begin{array}{c|c}\textbf{Tautology} \\ \hline \text{Conjunctive Simplification} & \begin{gathered} \left(p \wedge q\right) \rightarrow p \\ \left(p \wedge q\right) \rightarrow q \end{gathered} \\ \hline \text{Contradiction} & \neg \left(p \wedge \neg p\right) \\ \hline \text{Contrapositive} & \left(p \rightarrow q\right) \leftrightarrow \left(\neg q \rightarrow \neg p\right) \\ \hline \text{Disjunctive Addition} & \begin{gathered} p \rightarrow \left(p \vee q\right) \\ q \rightarrow \left(p \vee q\right) \end{gathered} \\ \hline \text{Disjunctive Syllogism} & \begin{gathered} \left(\left(p \vee q\right) \wedge \neg q\right) \rightarrow p \\ \left(\left(p \vee q\right) \wedge \neg p\right) \rightarrow q \end{gathered} \\ \hline \text{Hypothetical Syllogism} & \left(\left(p \rightarrow q\right) \wedge \left(q \rightarrow r\right)\right) \rightarrow \left(p \rightarrow r\right) \\ \hline \text{Modus Ponens} & \left(p \wedge \left(p \rightarrow q\right)\right) \rightarrow q \\ \hline \text{Modus Tollens} & \left(\neg q \wedge \left(p \rightarrow q\right)\right) \rightarrow \neg p \end{array}$$
asked 12 days ago
active 12 days ago