Is this a valid inference in propositional logic?

Question

As part of a proof I have $\vdash p \to (q \to r)$. Can I from this infer $\vdash (p \wedge q) \to r$? It "feels" correct, but I can't seem to find a proof.

Thanks for any help.

Answer 1

$\def\fitch#1#2{~\begin{array}{|l}#1\\\hline #2\end{array}}$

$$\fitch{p\to (q\to r)}{\fitch{p\wedge q}{q\\p\\q\to r\\r}\\(p\wedge q)\to r}\qquad\fitch{(p\wedge q)\to r}{\fitch{p}{\fitch{q}{p\wedge q\\ r}\\q\to r}\\p\to (q\to r)}$$

Answer 2

You can.

• Propositions-as-types proof: if you have a function $f: P \to (Q \to R)$, you can define a function $f' : P \times Q \to R$ by $f'(p, q) := f(p)(q)$.
• Valuations proof: just do the truth table. $p \to (q \to r)$ is false iff $p$ is true but $q \to r$ is false: that is, iff $p$ is true, $q$ is true, $r$ is false. On the other hand, $p \wedge q \to r$ is false iff $p \wedge q$ is true but $r$ is false; and that's obviously the same.
• Deduction theorem: the first one is $p, q \vdash r$; the second one is $p \wedge q \vdash r$. Those are manifestly the same.

Answer 3

By applying appropriate theorems to convert the LHS to RHS:

In this case:
$p→(q→r)$ is the LHS
$p→(q→r)$ is the RHS

Formulas Required:
$a→b ≡ ¬a∨b$
$¬(a∧b) ≡ ¬a∨¬b$

Steps:
$p→(q→r) ≡ p→(¬q∨r) ≡ ¬p∨(¬q∨r) ≡ (¬p∨¬q)∨r ≡ ¬(p∧q)∨r ≡ (p∧q)→r$