Using the laws of logic prove that $ [\neg q \land (p \rightarrow q)] \rightarrow \neg p$ is a tautology

Question

Could someone please tell me if I am correct and if I am not, tell me where I went wrong?

Using the laws of logic prove that $ [\neg q \land (p \rightarrow q)] \rightarrow \neg p$ is a tautology.

First I used the Implication law $(p \rightarrow q) \equiv (\neg p \vee q)$ to show that $$[\neg q \land (p \rightarrow q)] \equiv [\neg q \land (\neg p \vee q)]$$

Then I "factored" (?) the $\neg$ out and had $$ \neg [q \vee (p \land \neg q)] $$

And since $(p \land \neg q)$ denotes to "$p$ but not $q4" then I assumed I could leave $q$ out, leaving me with $$ ¬[q∨(p)] $$

Which is

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

Which says "not $q$ and not $p$".

Since it is "not $p$", does that mean that

$$ \equiv (\neg q \land \neg p) \rightarrow \neg p $$

and prove that $ [\neg q \land (p \rightarrow q)] \rightarrow \neg p$ is a tautology?

Answer 1

As Henning notes, you really do need to specify what laws are at your disposal. With that in mind, here is one approach (assuming you are able to use what I use): \begin{align} [\neg q\land(p\to q)]\to\neg p&\equiv\neg[\neg q\land(\neg p\lor q)]\lor\neg p\tag{material implication}\\[1em]&\equiv [q\lor(p\land\neg q)]\lor\neg p\tag{De Morgan}\\[1em] &= (q\lor\neg p)\lor(p\land\neg q)\tag{associativity}\\[1em] &\equiv \neg(p\land\neg q)\lor(p\land\neg q)\tag{De Morgan}\\[1em] &\equiv \neg M\lor M\tag{let $M\equiv p\land\neg q$}\\[1em] &\equiv \mathbf{T}\tag{negation}. \end{align}

Answer 2

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

$= -(-q\wedge(p\rightarrow q)) \vee -p$

$= (--q\vee-(-p\vee q))\vee -p$

$= (--q \vee (--p \wedge -q)) \vee -p$

$=(q \vee (p \wedge -q)) \vee -p$

$=((q \vee p) \wedge (q \vee -q)) \vee -p$

$=((q\vee p)\wedge 1) \vee (-p)$

$=(q \vee p) \vee (-p) = q \vee (p \vee -p) = q \vee 1 = 1$

Hence it is a tautology. I used the rules $--a = a$ and $a \wedge 1= a$, $a \vee 1 = 1$

Answer 3

$[\neg q \land (p \rightarrow q)] \rightarrow \neg p$

$\equiv [\neg q \land (\neg p \vee q)] \rightarrow \neg p$

$\equiv [q \vee \neg(\neg p \vee q)] \vee \neg p$

$\equiv q \vee (p \land \neg q) \vee \neg p$

$\equiv [(q \vee p) \land (q \vee \neg q)] \vee \neg p$

$\equiv q \vee 1$

$\equiv 1$