Negation: It means the opposite of the original statement. If p is a statement, then the negation of p is denoted by ~p and read as ‘it is not the case that p.’ So, if p is true then ~ p is false and vice versa.
Example: If statement p is Paris is in France, then ~ p is ‘Paris is not in France’.
| p | ~ p |
| T | F |
| F | T |
2. Conjunction: It means Anding of two statements. If p, q are two statements, then “p and q” is a compound statement, denoted by p ∧ q and referred as the conjunction of p and q. The conjunction of p and q is true only when both p and q are true. Otherwise, it is false.
| p | q | p ∧ q |
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |
3. Disjunction: It means Oring of two statements. If p, q are two statements, then “p or q” is a compound statement, denoted by p ∨ q and referred to as the disjunction of p and q. The disjunction of p and q is true whenever at least one of the two statements is true, and it is false only when both p and q are false.
| p | q | p ∨ q |
| T | T | T |
| T | F | T |
| F | T | T |
| F | F | F |
4. Implication / if-then (⟶): An implication p⟶q is the proposition “if p, then q.” It is false if p is true and q is false. The rest cases are true.
| p | q | p ⟶ q |
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | F |
5. If and Only If (↔): p ↔ q is bi-conditional logical connective which is true when p and q are same, i.e., both are false or both are true.
| p | q | p ↔ q |
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | T |
Derived Connectors
1. NAND: It means negation after ANDing of two statements. Assume p and q be two propositions. Nanding of pand q to be a proposition which is false when both p and q are true, otherwise true. It is denoted by p ↑ q.
| p | q | p ∨ q |
| T | T | F |
| T | F | T |
| F | T | T |
| F | F | T |
2. NOR or Joint Denial: It means negation after ORing of two statements. Assume p and q be two propositions. NORing of p and q to be a proposition which is true when both p and q are false, otherwise false. It is denoted by p ↑ q.
| p | q | p ↓ q |
| T | T | F |
| T | F | F |
| F | T | F |
| F | F | T |
3. XOR: Assume p and q be two propositions. XORing of p and q is true if p is true or q is true but not both and vice-versa. It is denoted by p ⨁ q.
| p | q | p ⨁ q |
| T | T | F |
| T | F | T |
| F | T | T |
| F | F | F |
Example1: Prove that X ⨁ Y ≅ (X ∧∼Y)∨(∼X∧Y).
Solution: Construct the truth table for both the propositions.
| X | Y | X⨁Y | ∼Y | ∼X | X ∧∼Y | ∼X∧Y | (X ∧∼Y)∨(∼X∧Y) |
| T | T | F | F | F | F | F | F |
| T | F | T | T | F | T | F | T |
| F | T | T | F | T | F | T | T |
| F | F | F | T | T | F | F | F |
As the truth table for both the proposition is the same.
- X ⨁ Y ≅ (X ∧∼Y)∨(∼X∧Y). Hence Proved.
Example2: Show that (p ⨁q) ∨(p↓q) is equivalent to p ↑ q.
Solution: Construct the truth table for both the propositions.
| p | q | p⨁q | (p↓q) | (p⨁q)∨ (p↓q) | p ↑ q |
| T | T | F | F | F | F |
| T | F | T | F | T | T |
| F | T | T | F | T | T |
| F | F | F | T | T | T |
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