Prove (p → ¬q) is equivalent to ¬(p ∧ q)

Question

I need to prove the above sequent using natural deduction. I did the first half already i.e. I proved $(p\rightarrow\neg q)\rightarrow \neg (p \wedge q)$, but I'm stuck on where to start for the reverse i.e. proving $\neg (p \wedge q) \rightarrow (p\rightarrow\neg q)$. I figured I would start by assuming $\neg (p \rightarrow \neg q)$ and then working towards a contradiction, but I'm still at a dead end. Can someone point me in the right direction? Thanks.

Answer 1

Can we use that $a \to b$ is same as $\neg a \lor b$? If yes then: $$\neg (p \wedge q) \rightarrow (p\rightarrow\neg q) $$ is same as $$(\neg p \lor \neg p) \rightarrow (\neg p \lor \neg q) $$

Answer 2

As a hint, note that $\neg p$ means $p \rightarrow \hbox{False}$. (In some logics, this is the definition of negation.)

Therefore $\neg (p \wedge q) \leftrightarrow (p \rightarrow \neg q)$ means: $$\left( \left(p \wedge q\right) \rightarrow \hbox{False} \right)\leftrightarrow \left(p \rightarrow q \rightarrow \hbox{False}\right)$$

This is just Currying.

Answer 3

One of the ways is this:

LHS

We already know that

(→¬) = (¬ + ¬)

RHS

By demorgan's law,

¬(∧) = (¬ + ¬)

Since LHS and RHS are same, so they are equivalent.

Answer 4

Here is a proof tree in natural deduction of the wanted implication, using $\Gamma = \{\neg(p\land q), p , q\}$:

$$\dfrac{ \dfrac{ \dfrac{ \dfrac{ \dfrac{ \dfrac{}{\Gamma\vdash p}\text{ax} \quad \dfrac{}{\Gamma\vdash q}\text{ax} }{\Gamma\vdash p\land q}\land_i \quad\dfrac{}{\Gamma\vdash \neg(p\land q)}\text{ax} }{\Gamma\vdash \bot}\neg_e }{\neg(p\land q), p\vdash \neg q}\neg_i }{\neg(p\land q)\vdash p\rightarrow \neg q}\rightarrow_i }{\vdash \neg(p\land q)\rightarrow (p \rightarrow \neg q)}\rightarrow_i$$

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