Back to the question :

Since Boolean variables can only have 2 different values, proving a Boolean algebra equality can always be done exhaustively. Or am I wrong ?

Yes. you_are_right or not(you are right) = true. That's basic binary logic.

Lame joke aside,

• Boolean algebras are (usually defined as) complemented distributive lattices.
• the two-element algebra (true, false, and, or, not, whatever) is one of them
• it has an interesting property:

identities that hold in the specific two-element algebra are exactly the same as the identities that hold for ALL boolean algebras. No more, no less.

So proving equalities with truth tables is a perfectly correct method, although a somehow boring activity.

The point is you want to teach them something else: using identities to prove other identities, or more generally turn terms into equivalent ones using algebraic manipulation of expressions.

The truth-tables method has an obvious adavantage for students: they don't have to think, just compute. When you use algebra, you have to plan an objective (the kind of expression you want to get) and find how you can get there step by step. It is sometimes tricky, like introducing instances of X or not(X) in a conjunction. It requires imagination, and taking decisions with try-and-error tentatives. Most students hate this.

See remark on https://en.wikipedia.org/wiki/Boolean_algebra_(structure)#Examples

The two-element Boolean algebra is also important in the general theory of Boolean algebras, because an equation involving several variables is generally true in all Boolean algebras if and only if it is true in the two-element Boolean algebra (which can be checked by a trivial brute force algorithm for small numbers of variables). This can for example be used to show that the following laws (Consensus theorems) are generally valid in all Boolean algebras

(a ∨ b) ∧ (¬a ∨ c) ∧ (b ∨ c) ≡ (a ∨ b) ∧ (¬a ∨ c)
(a ∧ b) ∨ (¬a ∧ c) ∨ (b ∧ c) ≡ (a ∧ b) ∨ (¬a ∧ c)

Back to the question :

Since Boolean variables can only have 2 different values, proving a Boolean algebra equality can always be done exhaustively. Or am I wrong ?

Yes. you_are_right or not(you are right) = true. That's basic binary logic.

Lame joke aside,

• Boolean algebras are (usually defined as) complemented distributive lattices.
• the two-element algebra (true, false, and, or, not, whatever) is one of them
• it has an interesting property:

identities that hold in the specific two-element algebra are exactly the same as the identities that hold for ALL boolean algebras. No more, no less.

So proving equalities with truth tables is a perfectly correct method, although a somehow boring activity.

The point is you want to teach them something else: using identities to prove other identities, or more generally turn terms into equivalent ones using algebraic manipulation of expressions.

The truth-tables method has an obvious adavantage for students: they don't have to think, just compute. When you use algebra, you have to chose a target (the kind of expression you want to get) and find how you can get there step by step. It is sometimes tricky, like introducing instances of X or not(X) in a conjunction. It requires imagination, and taking decisions with try-and-error tentatives. Most students hate this.

See remark on https://en.wikipedia.org/wiki/Boolean_algebra_(structure)#Examples

The two-element Boolean algebra is also important in the general theory of Boolean algebras, because an equation involving several variables is generally true in all Boolean algebras if and only if it is true in the two-element Boolean algebra (which can be checked by a trivial brute force algorithm for small numbers of variables). This can for example be used to show that the following laws (Consensus theorems) are generally valid in all Boolean algebras

(a ∨ b) ∧ (¬a ∨ c) ∧ (b ∨ c) ≡ (a ∨ b) ∧ (¬a ∨ c)
(a ∧ b) ∨ (¬a ∧ c) ∨ (b ∧ c) ≡ (a ∧ b) ∨ (¬a ∧ c)