Prove propositional formula is a theorem

Question

I need to show this formula is a theorem of propositional calculus. I tried assuming antecedent and proving consequent but didn't work for this proof. Do I need to show it is equivalent to true? How should I start?

$(p \rightarrow q) \rightarrow (p \wedge r \rightarrow q \wedge r)$

Background:

In my class, there is no mention to it but I think this form of logic it's called Equational Logic. This is a sample demonstration.

Answer 1

If we want to use the proof system of David Gries & Fred Schneider, A Logical Approach to Discrete Math (Springer, 1993), we can mimick the proof of (4.2), page 70 :

We begin with the consequent : $(p ∧ r \Rightarrow q ∧ r)$, since it has more structure, and transform it into the antecedent : $p \Rightarrow q$. Keeping in mind the goal, the first step is to eliminate the implication. Any of the four "definitions" of implication (3.57), (3.59), (3.60), and (3.61) could be used for this.

We have to use (3.60) : $(\alpha \Rightarrow \beta) \equiv (\alpha \land \beta \equiv \alpha)$ to get :

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

Using Idempotency, Associativity and Simmetry we get :

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

Now we use (3.62) : $\alpha \Rightarrow (\beta \equiv \gamma) \equiv (\alpha \land \beta) \equiv (\alpha \land \gamma)$, to get :

$$r \Rightarrow ((p \land q) \equiv p).$$

Now we use (4.1) : $\alpha \Rightarrow (\beta \Rightarrow \alpha )$ to derive (this is not an equivalence but $\Leftarrow$) :

$$(p \land q) \equiv p.$$

The last step uses again (3.60) to conclude with :

$$p \Rightarrow q.$$

In conclusion, we have :

$(p ∧ r \Rightarrow q ∧ r) \Leftarrow (p \Rightarrow q):$