determine whether ¬(p∨¬(p→q))∨p is a tautology by using laws of logic.

Question

¬(p ∨ ¬(p→q)) ∨ p ≡ ¬(p ∨ ¬(¬p ∨ q) ∨p (implication rule)
≡ ¬(p ∨(p ∨ ¬q) ∨p (double negation rule)
≡ ¬(p ∨ p) ∨ ¬q) ∨p (associative rule)
≡ ¬((p ∨ ¬q)∨p (Idempotent rule)
≡ (¬p ∨ q) ∨p(double negation rule) *I put the negation into p,do i need to change the sign?
≡ p v (¬p v q)
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Hi, I am new to discrete mathematics and i have no idea whether this is correct or not. Can anyone help me with this?

Answer 1

\begin{align}\neg(p \lor \neg(p\to q)) \lor p &\equiv \neg(p ∨ \neg(\neg p \lor q) \color{red})\lor p \text{(implication rule) }\\ &\equiv \neg(p \lor(p \color{red}{\land} ¬q)\color{red}) ∨p \text{(double negation rule and De Morgan) }\\ &\equiv \neg(p ) \lor p \\ &\equiv T \end{align}

In your second and fourth step, remember that when you bring in the negation, use De Morgan's rule.

Answer 2

As Mauro noted, you have an issue in your second line--you actually misapply DeMorgan's law (the disjunction should change to a conjunction), but you really do not need to use it there.

You are also consistently missing a right-hand parenthesis throughout.

Consider the following (I'll leave it to you to label the deductions.): $$ \begin{align*} \neg[p\lor\neg(p\to q)]\lor p &\equiv [\neg p\land(\neg p\lor q)]\lor p\\[1em] &\equiv (\neg p\land\neg p)\lor(\neg p\land q)\lor p\\[1em] &\equiv (\neg p\lor p)\lor(\neg p\land q)\\[1em] &\equiv \mathbf{T}\lor(\neg p\land q)\\[1em] &\equiv\mathbf{T}. \end{align*} $$