Propositional logic: Proof question (p∧q)→r⊢(p→q)→r

Question

Am I correct to assume that there is no proof for $$(p∧q)→r ⊢ (p→q)→r$$ I´ve spent hours trying to figure it out, by now I suspect there might have been a mistake in the exercise. I have been able to proof $$(p∧q)→r ⊢ p→(q→r)$$ (using Fitch notation), so it seems unlikely to me that $$(p∧q)→r ⊢ (p→q)→r$$ is valid as well. I´m quite new to propositional logic, so I just wanted to ask whether my reasoning is sound!

Answer 1

That's correct. There is no proof, because $p = F$ and $r = F$ is a counterexample to the argument.

And by the way, just because you have

$$(p \land q) \rightarrow r \vdash p \rightarrow (q \rightarrow r)$$

does not mean that you would not have

$$(p \land q) \rightarrow r \vdash (p \rightarrow q) \rightarrow r$$

even if $p \rightarrow (q \rightarrow r)$ is not the same as $(p \rightarrow q) \rightarrow r$.

For example: we have that $p \land q \vdash p$ ... but does that mean that therefore we don't have $p \land q \vdash q$ (because $p$ is not $q$)? Clearly not.

Answer 2

There is no proof because the tableau of $\lnot (((p\land q)\to r)\to ((p\to q)\to r)))$ is not closed.