How to simplify $(p \to (p \land q)) \to q$ without using implication?

Question

This is what I have done:

$(p \to (p\land q)) \to q$

= ~(~p ∨ (p∧q)) ∨ q

= (p ∧ ~(p∧q)) ∨ q

= (p v q) ∧ ~(p∧q)

= ~(p∧q)

= ~p v ~q

May I ask if my steps thus far are correct? If not, may I ask which step have I made a mistake? Thank you.

Answer 1

In the third step: $$ (p \land \lnot (p \land q)) \lor q \;\equiv\; (p \lor q) \land \color{red}(\lnot (p \land q) \color{red}{\lor q)} $$

Answer 2

Note that below you'll find two ways of simplifying the given proposition, each of which simplify to $(p\lor q)$:

(1) I'll start with your third step: $$ (p\land \lnot (p\land q)) \lor q\tag 3$$

For the fourth step, we use DeMorgan's again, this time on the clause $\lnot (p\land q)\equiv (\lnot p \lor \lnot q)$:

$$(p \land \lnot (p\land q)) \lor q \equiv (p\land (\lnot p \lor \lnot q))\lor q\tag{5, DeMorgan's}$$

$$\equiv \big(\underbrace{(p\land \lnot p)}_{\large \text{False}} \lor (p \land \lnot q)\big) \lor q\tag{distribution}$$

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

$$\equiv (p \lor q) \land \underbrace{(\lnot q\lor q)}_{\large \text{True}} \tag{distribution}$$

$$\equiv p \lor q $$

---

(2) Another approach just switches the order in which we distribute, and the order we use DeMorgan's

Note that you have from step $3\to 4$ the following:

$$(p \land \lnot (p\land q)) \lor q\tag 3$$

$$\equiv (p \lor q) \land (\lnot(p\land q)\ldots)\tag{4,$\color{red}{\Large \checkmark}$, erroneous distribution}$$

Rather, we must have, instead,

$$\equiv (p\lor q) \land (\lnot (p \land q) \lor q)\tag {5, Distribution from 3}$$

$$\equiv (p \lor q) \land \underbrace{(\lnot p\lor \underbrace{\lnot q \lor q}_\text{True})}_\text{True}\tag {6, DeMorgan's}$$

$$\equiv (p \lor q) \land \text{True}\tag 7$$

$$\equiv p \lor q\tag 8$$

$\color{red}{\Large \checkmark}$ The problem is that you left out in $(4)$ what should replace ($\;\cdots\;$) above, and that is: $\;\lor q$.