p ∧ (q ∧ r ) ≡ (p ∧ q) ∧ rp ∨ (q ∨ r ) ≡ (p ∨ q) ∨ r
To prove the given equations, we will use the laws of boolean algebra
To prove the given equations, we will use the laws of boolean algebra. Here’s the step-by-step breakdown:
1. p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r
This equation is applying the associative law of conjunction, which states that for any propositions p, q, and r, (p ∧ q) ∧ r is logically equivalent to p ∧ (q ∧ r).
To prove the equivalence, we’ll use a truth table:
p | q | r | q ∧ r | p ∧ (q ∧ r) | p ∧ q | (p ∧ q) ∧ r
—————————————————–
T | T | T | T | T | T | T
T | T | F | F | F | T | F
T | F | T | F | F | F | F
T | F | F | F | F | F | F
F | T | T | T | F | F | F
F | T | F | F | F | F | F
F | F | T | F | F | F | F
F | F | F | F | F | F | F
As seen in the truth table, the values of p ∧ (q ∧ r) and (p ∧ q) ∧ r are the same for all possible combinations of p, q, and r. Therefore, p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r.
2. p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r
This equation is applying the associative law of disjunction, which states that for any propositions p, q, and r, (p ∨ q) ∨ r is logically equivalent to p ∨ (q ∨ r).
Again, we’ll use a truth table to prove the equivalence:
p | q | r | q ∨ r | p ∨ (q ∨ r) | p ∨ q | (p ∨ q) ∨ r
—————————————————–
T | T | T | T | T | T | T
T | T | F | T | T | T | T
T | F | T | T | T | T | T
T | F | F | F | T | T | T
F | T | T | T | T | T | T
F | T | F | T | T | T | T
F | F | T | T | T | F | T
F | F | F | F | F | F | F
As shown in the truth table, the values of p ∨ (q ∨ r) and (p ∨ q) ∨ r are the same for all possible combinations of p, q, and r. Therefore, p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r.
Hence, we have successfully proved the given equations using the laws of boolean algebra.
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