## p ∧ (q ∧ r ) ≡ (p ∧ q) ∧ rp ∨ (q ∨ r ) ≡ (p ∨ q) ∨ r

### To prove the two given expressions:

Expression 1: p ∧ (q ∧ r ) ≡ (p ∧ q) ∧ r

Using the associative property of ∧ (logical AND), we can rearrange the parentheses:

p ∧ (q ∧ r ) ≡ (p ∧ r) ∧ q

Now let’s consider the truth table for logical AND:

p | q | r | q ∧ r | p ∧ (q ∧ r) | (p ∧ r) ∧ q

—————————————–

T | T | T | T | T | T

T | T | F | F | F | F

T | F | T | F | F | F

T | F | F | F | F | F

F | T | T | T | F | F

F | T | F | F | F | F

F | F | T | F | F | F

F | F | F | F | F | F

As we can see from the truth table, the values for both Expression 1 are always the same (either T or F) for all possible combinations of p, q, and r

To prove the two given expressions:

Expression 1: p ∧ (q ∧ r ) ≡ (p ∧ q) ∧ r

Using the associative property of ∧ (logical AND), we can rearrange the parentheses:

p ∧ (q ∧ r ) ≡ (p ∧ r) ∧ q

Now let’s consider the truth table for logical AND:

p | q | r | q ∧ r | p ∧ (q ∧ r) | (p ∧ r) ∧ q

—————————————–

T | T | T | T | T | T

T | T | F | F | F | F

T | F | T | F | F | F

T | F | F | F | F | F

F | T | T | T | F | F

F | T | F | F | F | F

F | F | T | F | F | F

F | F | F | F | F | F

As we can see from the truth table, the values for both Expression 1 are always the same (either T or F) for all possible combinations of p, q, and r. Therefore, we can conclude that Expression 1 is true.

Expression 2: p ∨ (q ∨ r ) ≡ (p ∨ q) ∨ r

Using the associative property of ∨ (logical OR), we can rearrange the parentheses:

p ∨ (q ∨ r ) ≡ (p ∨ r) ∨ q

Now let’s consider the truth table for logical OR:

p | q | r | q ∨ r | p ∨ (q ∨ r) | (p ∨ r) ∨ q

—————————————–

T | T | T | T | T | T

T | T | F | T | T | T

T | F | T | T | T | T

T | F | F | F | T | T

F | T | T | T | T | T

F | T | F | T | T | T

F | F | T | T | T | T

F | F | F | F | F | F

As we can see from the truth table, the values for both Expression 2 are always the same (either T or F) for all possible combinations of p, q, and r. Therefore, we can conclude that Expression 2 is true.

So, both Expression 1 and Expression 2 are true.

## More Answers:

Proving the Equivalence of ¬(¬p) and p in Mathematics: Step-by-Step AnalysisProving Logical Equivalences Using De Morgan’s Laws: A Step-by-Step Guide

Proving the Equality p ∧ q ≡ q ∧ p: A Truth Table Approach