Prove ((p→q)∧q) and q are equivalent using logic laws

Question

So far I have

((p→q)∧q)
≡ ((~p∨q)∧q)
≡ (~p∧q)∨(q∧q) $\; \; \; \; \;$(Distributive law)
≡ (~p∧q)∨q

But then I got stuck. And I'm not sure if what I've got so far is correct. Any help is greatly appreciated!

Answer 1

Use the Absorption laws $$a \land (a \lor b) \equiv a$$ $$a \lor (a \land b) \equiv a$$ We can easily verify this.

Answer 2

I offer a proof of the syntactic equivalence $(p \to q) \wedge q \dashv \vdash q$ via natural deduction using only conditional proof and intro/elimination rules for conjunction:

$(p \to q) \wedge q \vdash q$

$ \begin{array}{11111} \{1\} & 1. & (p \to q) \wedge q & \text{premise} \\ \{1\} & 2. & q& \text{1 Conjunction Elimination} & \square\\ \end{array} $

$ q \vdash (p \to q) \wedge q$

$ \begin{array}{11111} \{1\} & 1. & q & \text{premise} \\ \{2\} & 2. & p & \text{Assumption for Conditional Proof} \\ \{1,2\} & 3. & p \wedge q & \text{1,2 Conjunction Introduction} \\ \{1,2\} & 4. & q& \text{3 Conjunction Elimination} \\ \{1\} & 5. & p \to q & \text{2,4 Conditional Proof} \\ \{1\} & 6. & (p \to q) \wedge q& \text{1,5 Conjunction Introduction} & \square \\ \end{array} $