I was trying to simplify this propositional form:
$$ \{(p \lor q) \land [¬q \lor (p \lor t)] \land (¬t \lor p) \} \lor (p \land ¬t)$$
but ended up with: $$\ (q \lor p) \lor (p \land ¬t)$$
My steps were:
$$ \{(q \lor p) \land [p \lor (¬q \lor t)] \land (¬t \lor p) \} \lor (p \land ¬t)$$
$$ \{q \lor [p \land [p \lor (¬q \lor t)]] \land (¬t \lor p) \} \lor (p \land ¬t)$$
$$\{ q \lor [p \land (¬t \lor p)]\} \lor (p \land ¬t)$$ $$\{ q \lor [p \land (p \lor ¬t)]\} \lor (p \land ¬t)$$ $$\ (q \lor p) \lor (p \land ¬t)$$
I need help to further simplifying this expression, any help will be really appreciated, Thanks in advance!
First step, distribute rather than associate $(a\vee b)\wedge c = (a\wedge c)\vee(b\wedge c) \quad\neq a\vee(b \wedge c)$
$${ \{(q \lor p) \land [p \lor (¬q \lor t)] \land (¬t \lor p) \} \lor (p \land ¬t)\\[1ex] \{[(q \land [p \lor (¬q \lor t)])\lor (p \land [p \lor (¬q \lor t)])] \land (¬t \lor p) \} \lor (p \land ¬t) }$$
However, I would not go that way. Rather I suggest the first step should be to distribute out the common disjunct. Thence you may associate over the same operator.
$${ \{(q \lor p) \land [p \lor (¬q \lor t)] \land (¬t \lor p) \} \lor (p \land ¬t)\\[1ex] (\{q \land (¬q \lor t) \land ¬t \} \lor p) \lor (p \land ¬t) \\[1ex] \{q \land (¬q \lor t) \land ¬t \} \lor (p \lor (p \land ¬t)) \\[1ex]\vdots }$$
Now, proceed from there.