I keep trying to prove the propositional logic theorem:
$$\vdash (p \rightarrow q) \wedge (p \wedge q \rightarrow r) \rightarrow (p \rightarrow r)$$
using the theorem of deduction and its reverse (and obviously using the axioms and modus ponens), but I get nowhere. I've tried every possibility that I could come up with, I'm starting to lose my mind. The fact that I have $2$ conjunctions within the formula ruins everything. So I know I need to use the reverse of the theorem of deduction to simplify things (so we successively move the left sides of implications to the left of the deduction symbol $\vdash$):
$$ (p \rightarrow q) \wedge (p \wedge q \rightarrow r) \vdash p \rightarrow r$$
$$(p \rightarrow q) \wedge (p \wedge q \rightarrow r), p \vdash r$$
And from here I keep getting stuck. I tried to transform that conjunction into an implication using:
$$U \wedge V \equiv \neg (\neg U \lor \neg V) \equiv \neg(U \rightarrow \neg V)$$
And I use this twice, for both conjunctions. But the formula gets so complicated that I can't see anything and there seems to be no place to use modus ponens on. And if I use the above on just one conjunction I still don't get anything that I can use.
Best Answer
Use conditional proof inference twice. Statement $$ (p \rightarrow q) \wedge (p \wedge q \rightarrow r) \vdash p \rightarrow r$$
is equivalent to $$ (p \rightarrow q),(p \wedge q \rightarrow r) \vdash p \rightarrow r$$
and then $$(p \rightarrow q),(p \wedge q \rightarrow r),p \vdash r$$
which is obviously valid statment.