The Identity Law can be verified with this truth table. The fact that \((p \land 1)\leftrightarrow p\) is a tautology serves as a valid proof.
\(p\) | \(1\) | \(p\land 1 \) | \((p\land 1)\leftrightarrow p\) |
0 | 1 | 0 | 1 |
1 | 1 | 1 | 1 |
\(p\) | \(1\) | \(p\land 1 \) | \((p\land 1)\leftrightarrow p\) |
0 | 1 | 0 | 1 |
1 | 1 | 1 | 1 |
Commutative Laws | ||
\(p \lor q\Leftrightarrow q\lor p\) | \(p \land q\Leftrightarrow q \land p\) | |
Associative Laws | ||
\((p \lor q) \lor r \Leftrightarrow p \lor (q \lor r)\) | (\(p \land q) \land r\Leftrightarrow p \land (q \land r)\) | |
Distributive Laws | ||
\(p \land (q \lor r) \Leftrightarrow (p \land q ) \lor (p \land r)\) | \(p \lor (q \land r) \Leftrightarrow (p \lor q) \land (p \lor r)\) | |
Identity Laws | ||
\(p \lor 0\Leftrightarrow p\) | \(p \land 1 \Leftrightarrow p\) | |
Negation Laws | ||
\(p\land \neg p\Leftrightarrow 0\) | \(p\lor \neg p\Leftrightarrow 1\) | |
Idempotent Laws | ||
\(p \lor p \Leftrightarrow p\) | \(p\land p \Leftrightarrow p\) | |
Null Laws | ||
\(p \land 0 \Leftrightarrow 0\) | \(p \lor 1 \Leftrightarrow 1\) | |
Absorption Laws | ||
\(p \land (p\lor q)\Leftrightarrow p\) | \(p \lor (p \land q) \Leftrightarrow p\) | |
DeMorgan’s Laws | ||
\(\neg (p \lor q) \Leftrightarrow (\neg p) \land (\neg q)\) | \(\neg (p \land q) \Leftrightarrow (\neg p) \lor (\neg q)\) | |
Involution Law | ||
\(\neg (\neg p)\Leftrightarrow p\) |
Detachment (AKA Modus Ponens) | \((p \rightarrow q) \land p\Rightarrow q\) |
Indirect Reasoning (AKA Modus Tollens) | \((p \to q) \land \neg q \Rightarrow \neg p\) |
Disjunctive Addition | \(p\Rightarrow (p\lor q)\) |
Conjunctive Simplification | \((p \land q) \Rightarrow p\) and \((p \land q) \Rightarrow q\) |
Disjunctive Simplification | \((p \lor q) \land \neg p \Rightarrow q\) and \((p \lor q) \land \neg q\Rightarrow p\) |
Chain Rule | \((p \to q) \land ( q \rightarrow r) \Rightarrow (p\to r)\) |
Conditional Equivalence | \(p \rightarrow q \Leftrightarrow \neg p \lor q\) |
Biconditional Equivalences | \((p \leftrightarrow q) \Leftrightarrow (p\rightarrow q) \land (q \rightarrow p)\Leftrightarrow (p \land q) \lor (\neg p \land \neg q)\) |
Contrapositive | \((p\to q) \Leftrightarrow (\neg q \to \neg p)\) |