Discrete mathematics Logic Proof

Question

I'm stuck with these problems...

 1.Show that B∧C,(B↔C)→H∨G tautologically implies(⇒) G∨H ? 
 2.Use Rule Conditional Proof (C.P) show that P→(Q→R) , Q→(R→S) tautologically implies(⇒) P→(Q→S)
 3.Using Proof by Contradiction, show that P→¬S is valid. (P→ (Q∧R))∧ (Q→¬P) ∧ (S→¬R) ∧P

My attempt to answer the first question .

 i) B∧C Rule P 
 ii)B Rule T (i) Simplification 
 iii)C Rule T (i) 


after this i should arrive at (B↔C) . How should i do this ?

My attempt to answer the second question .

i)P Rule P(Assumed)
ii) P→(Q→R) Rule P
iii) Q→R Rule T (i)(ii) Modus Ponens 
i should get Q from this . How should i do this ? 
note: Can we assume One more premise Q, to get the answer is it possible . 

My attempt to answer the third question .

Given Premises : (P→(Q∧R)),(Q→¬P),(S→¬R)∧P
Required Conclusion : P→¬S


i)¬(P→¬S) Rule P ( Assumed)
ii) 
What should i do next ?

Can anyone explain me how to do these problems . I don't know the exact symbol used for tautological implication .

Answer 1

I assume that you are using natural deduction:

(1) Prove that $B\land C,(B \leftrightarrow C) \to (H \lor G) \vdash G \lor H$

Proof sketch: It is not hard to show that $B∧C \to (B↔C)$ (e.g. here). Now you can derive $H \vee G$ by modus ponens and $G \vee H$ from the commutativity of disjuction.

(2) Prove that $P\to (Q \to R) , Q \to (R \to S) \vdash P \to (Q \to S)$

Hint: What happens if you assume $P$ as hypothesis, and then assume $Q$ afterwards?

(3) Prove that $P \to (Q\land R),Q\to \neg P,(S\to \neg R) \land P \vdash P→¬S$

Hint: Since you seem to require a proof by contradiction, assume $\neg (P→¬S)$. By the De Morgan laws, this is the same as $P \wedge S$. By conjunction elimination, we have derivations of $P$ and $S$.

Let's look at our premises once again, is there any way we can get a contradiction, say, $R \wedge \neg R$?