I guess on stackoverflow you'd get closed for asking homework questions but as I couldn't find rules against that (yet, correct me if I'm wrong), let's do this logical sudoku real quick:
Assuming that this operator ⊢ stand for equality of tautology or something like that https://en.wikipedia.org/wiki/Logical_connective
Now validity of a statement means that if the premises were true the conclusion must also be true. So it's essentially:
Premise1∧Premise2 → Conclusion.
((Q → ¬P) ∧ (¬Q → ¬P)) → ¬P
Now you can apply a Simplification of disjunctive antecedents (* see proof of why that can be applied further down in this answer)
((Q ∨ ¬Q) → ¬P) → ¬P
Now (Q ∨ ¬Q) is obviously true, so there's no harm done adding another true statement with an and due to the fact that True And Statement would still obviously have the same truth value as the statement itself.
(Q ∨ ¬Q) ∧ ((Q ∨ ¬Q) → ¬P) → ¬P
Now if we look closely at this we could see that his is modus ponens and therefore a tautology. Which can be made more clear by replacing (Q ∨ ¬Q) with idk a symbol R of the same truth value:
R ∧ (R → ¬P) → ¬P
Where as said R ∧ (R → ¬P) is ¬P due to MP, which concludes this proof.
And for the other one (that is now no longer part of the question, but was when this answer was written...):
(Q → ¬R), (¬P → R) ⊢ (¬P → ¬Q)
One can use the Contraposition A → B = ¬B → ¬A. So:
(Q → ¬R) ∧ (¬P → R) → (¬P → ¬Q)
Becomes:
(R → ¬Q) ∧ (¬P → R) → (¬P → ¬Q)
And from there it follows from the transitive property of the Material conditional that ¬P → ¬Q at which it again follows from itself which also concludes that it's valid.
So you should obviously look up what was done and why you are allowed to do that ;)
Update: A practice problem to study is still a home work ;)
Proof that you can rephrase SDA in terms of the copi rules (assuming those are your constraints)
Anyway well you need to rephrase these logical operations in terms of your equivalence and interference rules:
So for example you'd need to show that
(A ∨ B) → C is the same as A → C ∧ B → C :
So here you would apply imp to get:
¬(A ∨ B) ∨ C
From there you would apply De Morgan to the parenthesis:
(¬A ∧ ¬B) ∨ C
Assuming that the distributive law which isn't listed in the previous list but in this one:
https://en.wikipedia.org/wiki/Propositional_calculus
Is part of the derivation laws. You can write that as:
(¬A ∨ C) ∧ (¬B ∨ C)
Which if you can interpret as two imp statements connected with an AND:
(A → C) ∧ (B → C)
So the Simplification of disjunctive antecedents can indeed be applied even with the derivation laws of the propositional logic. Though you'd go the other way around, but as these are equivalence relations that shouldn't be a problem. I leave it to your practice to fit that into the right form or was that your problem?
For the Contraposition there's a nice simple proof on Wikipedia:
https://en.wikipedia.org/wiki/Contraposition#Simple_proof_by_definition_of_a_conditional
which should suffice the derivation rules.
And the transitive relation in proposition logic is apparently called Hypothetical syllogism which also has a prove written in the formal language.