Conjunction Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

medium
Construct the full truth table for p∧qp \land q and use it to show that conjunction is commutative: p∧q≡q∧pp \land q \equiv q \land p.

Solution

  1. 1
    List all rows: (T,T),(T,F),(F,T),(F,F)(T,T), (T,F), (F,T), (F,F).
  2. 2
    p∧qp \land q: T,F,F,FT, F, F, F.
  3. 3
    q∧pq \land p: also T,F,F,FT, F, F, F (same values, just pp and qq swapped).
  4. 4
    The columns are identical, confirming p∧q≡q∧pp \land q \equiv q \land p.

Answer

p∧q≡q∧pp \land q \equiv q \land p
Two formulas are logically equivalent when they have the same truth value in every row of the truth table. Commutativity of ∧\land follows directly from the symmetric definition: both components must be true.

About Conjunction

A conjunction P∧QP \wedge Q is a compound statement that is true if and only if both constituent statements PP and QQ are individually true.

Learn more about Conjunction →

More Conjunction Examples

Example 1 easy

Let [formula]: '[formula] is even' and [formula]: '[formula]'. Evaluate [formula], [formula], and [f

Example 3 easy

Determine the truth value of: (a) '[formula] and [formula]', (b) '[formula] and [formula]'.

Example 4 medium

Simplify: find all [formula] satisfying '[formula] and [formula]', and express as an interval.

← All Conjunction Examples Conjunction Concept