Is "(p AND q) OR r" logically equivalent to "p AND (q OR r)" ??

Question

In the context of discreet math / boolean algebra / logic, is "(p AND q) OR r" logically equivalent to "p AND (q OR r)"? I believe so, but my professor said:

They are not the same.  If p is true, q is false, and r is true, then the first expression is false:

(p and q) or r
becomes
(T and F) or T
which is
F or T
which is
F.


But the 2nd expression is p and (q or r) which is (T and (F or T)) = (T and T) = T.
These are different."

Is he correct? (Also, please excuse my formatting, this is my first question on Math)

Answer 1

The property that we're talking about is called Associativity. It's a property for a single operator $*$ defined as $(a*b)*c=a*(b*c)$. What you have is a $(a*b)+c$ which has 2 operators in it. For more information check out Associativity. Welcome to Mathematics.SE.

Answer 2

They are not the same.

p q r │ (p∧q)∨r  p∧(q∨r)
──────┼───────────────────
F F F │   F        F
F F T │   T        F     *
F T F │   F        F
F T T │   T        F     *
T F F │   F        F
T F T │   T        T
T T F │   T        T
T T T │   T        T

Answer 3

The professor gave a fine explanation. Two propositions are equivalent if and only if for any possible assignment of truth values to the variables, the propositions are both false, or both true, under the same assignment. Your professor provided a truth value assignment which renders one proposition true and the other, false. Hence the propositions cannot be equivalent.

Perhaps it will help to see how the propositions look after applying the distributive law that applies to each proposition you are comparing:

$$(p \land q) \lor r \equiv (p \lor r) \land (q \lor r)$$

$$(p \lor q) \land r \equiv (p \land r) \lor (q \land r)$$

Parentheses are crucial here, since they dictate what to evaluate first, and the order in which to evaluate. Without parentheses, the expressions are ambigous: we cannot conclude anything then. When you have mixed connectives involving, e.g., $\lor, \land$, we cannot use associativity to "mix and match" the variables as we please.