Is it true that $((p\land q \rightarrow r) \land \lnot(p \rightarrow r)) \rightarrow (q \rightarrow r)$?

Question

Is this a sound inference rule?

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

So far I've rewritten it to $$((p \rightarrow r) \land \lnot(p \rightarrow r)) \rightarrow (q \rightarrow r)$$

It looks similar to the syllogism rule, but I'm not sure. Any help?

Answer 1

@GEdgar's suggestion was spot on. If you write a truth tabe to consider $2^3 = 8$ possible truth-value assignments for $(p, q, r)$, you will see that your given expression is, in fact, a tautology, meaning that no matter what truth values you assign to $p, q$ and $r$, the propostion will always be true:

Hence, indeed, the implication is a valid and sound inference rule.

Answer 2

Not using your rules of inference, but mine...

[ 1]TO PROVE $((p \land q) \rightarrow r) \land \lnot(p \rightarrow r)) \rightarrow (q \rightarrow r)$
[ 2]$\quad$ ASSUME $((p \land q) \rightarrow r) \land \lnot(p \rightarrow r)$
[ 3]$\quad$ TO PROVE $q \rightarrow r$
[ 4]$\qquad$ ASSUME $q$
[ 5]$\qquad$ TO PROVE $r$
[ 6]$\qquad\qquad$ $\lnot(p \rightarrow r)\qquad$ by [2]
[ 7]$\qquad\qquad$ $\lnot(\lnot p \lor r)\qquad$ by [6], definition of $\rightarrow$
[ 8]$\qquad\qquad$ $p \land \lnot r\qquad$ by [7], DeMorgan
[ 9]$\qquad\qquad$ $p\qquad$ by [8]
[10]$\qquad\qquad$ $p \land q\qquad$ by [9],[4]
[11]$\qquad\qquad$ $(p\land q)\rightarrow r\qquad$ by [2]
[12]$\qquad\qquad$ $r\qquad$ by [11],[10], modus ponens, establishes [5]
[13]$\qquad$ $q\rightarrow r\qquad$ by [4]...[12], establishes [3]
[14]$ $ $((p \rightarrow r) \land \lnot(p \rightarrow r)) \rightarrow (q \rightarrow r)$, by [2]...[13], establishes [1]