Which of $\varphi$ or $\lnot \varphi$ can be expressed by using only the $\rightarrow$ connective?

Question

if we have:

$$\varphi = \lnot(p\land q\to r) $$

(original screenshot)

a) we can write $\varphi$ in equivalence just by using $\to$ connective.

b) we can write $\lnot\varphi$ in equivalence just by using $\to$ connective.

any experts can help me (a) or (b) is correct? why?

Answer 1

Hint:

$$x\lor y \equiv \lnot x\to y \\ x\land y \equiv \lnot(\lnot x \lor \lnot y) $$

Answer 2

$$\begin{align}\lnot \varphi & \equiv \lnot \Big(\lnot((p\land q)\rightarrow r)\Big) \\ & \equiv (p \land q)\rightarrow r\\ & \equiv p\rightarrow(q\rightarrow r) \end{align}$$

Answer 3

In general

$$A \rightarrow B = \lnot A \lor B$$

Also, notice that

$$\lnot(\lnot A \lor B) = A \land \lnot B ~\text{(De Morgan theorem)}$$

In your case, you have:

$$\varphi = \lnot((p \land q) \rightarrow r) = p \land q \land \lnot r $$

and

$$\lnot\varphi = (p \land q) \rightarrow r = \lnot p \lor \lnot q \lor r ~\text{(De Morgan theorem)}$$

In the second case, we can do the following:

$$\lnot\varphi = \lnot p \lor (\lnot q \lor r) = p \rightarrow (\lnot q \lor r) = p \rightarrow (q \rightarrow r) $$

Answer 4

Both are correct given that you have "0" as allowable in expressions since (p→0) is logically equivalent to ¬p.

¬φ is logically equivalent to (p→(q→r)).

¬(p∧q→r) is logically equivalent to ¬(¬(p→¬q)→r) which is logically equivalent

¬(¬(p→(q→0))→r) which is logically equivalent to

¬(((p→(q→0))→0)→r) which is logically equivalent to

((((p→(q→0))→0)→r)→0).