With this one you can use Taut Con (to implement MT, DS, DeM, if needed). Your main strategy is conditional proof/ →-Intro so you'll need a subproof.
premise 1: P → (Q → R) premise 2: P → (Q ∨ S) premise 3: P → ¬R prove: P → S
I tried this:
P → (Q → R) Premise
P → (Q ∨ S) Premise
P → ¬R Premise
Assume P Assumption
Assume Q Assumption
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P → R → Elimination (1)
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R → Elimination (4, 6)
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¬R → Elimination (3)
Q → ¬R → Introduction (5-8)
Assume S Assumption
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Q ∨ S → Elimination (2)
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¬R → Elimination (9, 10)
S ∨ Elimination (11)
P → S → Introduction (4-13)
I was expecting it all to be true but it was not.