Giving a formal proof of p ⇒(q ⇒ r) $\vdash$ (p ⇒ r)∨(q ⇒ r) using the rules of inference. I can prove this with semantic equivalences and truth tables but I'm struggling on the formal proof using rules of inference front.
Given its format I would assume it must finish on V-introduction which means using if-introduction at some point on p⇒r or q⇒r?
I've made 3 or 4 attempts now and I can't seem to get there!
Any advice or pointers in the right direction would be very much appreciated,
Thanks!
Edit: We've been using Γ ├ B to mean ‘B can be formally proved from the set of Γ of assumptions’.
Edit: The rules at my disposal are:


*

*&-I

*&-E

*V-I

*V-E

*⇔-I

*⇔-E

*⇒-I

*⇒-E

*$\lnot$-I

*$\lnot$-E

*true-I

*false-E 

 A: Your strategy of ending with a $\lor$ Introduction on either $p \rightarrow r$ or $q \rightarrow r$ is not going to work
If $p$ is true, but $q$ and $r$ is false, then $p \rightarrow (q \rightarrow r)$ is true, but $p \rightarrow r$ is false, so you will not be able to derive $p \rightarrow r$ by itself.
If $q$ is true, but $p$ and $r$ is false, then $p \rightarrow (q \rightarrow r)$ is true, but $q \rightarrow r$ is false, so you will not be able to derive $q \rightarrow r$ by itself.
Indeed, here is a general 'tip': if you ever have to prove a disjunction as your ultimate conclusion, then most likely you will not be able to prove either of then disjuncts by themselves!
So, what to do instead?  I would recommend either:


*

*Doing this by a proof by contradiction: if $(p \rightarrow r) \lor (q \rightarrow r)$ is false, then both $p \rightarrow r$ and $q \rightarrow r$ are false, hence $p$ and $q$ are true but $r$ is false, contradicting $p \rightarrow (q \rightarrow r)$


or:


*Start by proving $p \lor \neg p$. And then: if $p$, then $q \rightarrow r$ using Modus Ponens ($\rightarrow$ Elim I suspect) between $p$ and the premise. And if $\neg p$, then you can show $p \rightarrow r$. So, you can show $(p \rightarrow r) \lor (q \rightarrow r)$ in either case (so now you do $\lor$ Intro within each subproof .. that is something that will frequently occur in formal proofs). So, using proof by cases (presumably your $\lor$ Elim), you definitely have $(p \rightarrow r) \lor (q \rightarrow r)$

A: Since $(a \Rightarrow b) \Leftrightarrow (\lnot a \vee (a \wedge b)) \Leftrightarrow (\lnot a \vee b   ) $ for all $a,b$ by the material implication rule.
$$
[ p \Rightarrow (q \Rightarrow r) ] \Leftrightarrow [ \lnot p \vee (q \Rightarrow r) ] \Leftrightarrow [ \lnot p \vee ( \lnot q \vee r) ] 
\Leftrightarrow [ \lnot p \vee \lnot q \vee r ] 
$$
By the same rule,
$$
[ (p \Rightarrow r ) \vee  (q \Rightarrow r) ] \Leftrightarrow 
[ (\lnot p \vee r  ) \vee ( \lnot q \vee r) ] \Leftrightarrow 
[ \lnot p \vee  \lnot q \vee r] 
$$
Hence the result.
A: $$p \to(q \to r) \vdash (p \to r)\lor(q \to r)$$
LHS:$$p \to(q \to r)$$
$$\neg p \lor (q\to r)$$
$$\neg p \lor (\neg q\lor r)$$
$$\color{red}{\neg p\lor r\lor \neg q}$$
RHS: $$(p\to r)\lor(q\to r)$$
$$(\neg p \lor r)\lor(\neg q \lor r)$$
$$\color{red}{\neg p\lor r\lor \neg q}$$
