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
-
P → R → Elimination (1) -
R → Elimination (4, 6) -
¬R → Elimination (3) Q → ¬R → Introduction (5-8)
Assume S Assumption
-
Q ∨ S → Elimination (2) -
¬R → Elimination (9, 10) S ∨ Elimination (11)
P → S → Introduction (4-13)
I was expecting it all to be true but it was not.