# Assuming p $\to$ (q OR r) and p $\to$ (q OR NOT r), prove p $\to$ q

So I've been trying to figure out how to prove this statement:

Assuming p $\to$ (q OR r) and p $\to$ (q OR NOT r), prove p $\to$ q.

This is about as far as I got:

1. p → (q OR r) // Given
2. p → (q OR NOT r) // Given
3. p // Given (I assume that p must exist in order to prove that p → q)
4. q OR r // 1, 3, modus ponens
5. q OR NOT r // 2, 3, modus ponens
6. (q OR r) AND (q OR NOT r) // 4, 5, conjunction

I'm not sure if I'm on the right track, and even if I am, I don't know how to proceed... Could I get a hint? Thanks in advance!

• \to gives a one lined arrow, \implies gives a two-lined arrow (remember that you have to surround MathJax with $signs to work -- also MathJax is just web-enabled LateX) Commented Oct 8, 2016 at 19:17 ## 1 Answer So far so good....you've arrived at:$(6)\quad (q \lor r)\land (q\lor \lnot r)\tag{correct}$Your next step, by using the distributive property of "or" over$\land\$, we have

$$(7) \quad q\lor (r \land \lnot r)\tag{from (6) Distributive property}$$

$$(8)\quad q \lor (F)\tag{from (7) contradiction}$$

$$(9) \quad q\tag{from (8)}$$ $$(10)\quad p\rightarrow q \;\;(3, 9)\tag{conditional introduction}$$