Supposedly the conclusion follows but I don't see how, since if we imagine $S$ as false but $P$ as true, then things seem to check out. (Classical predicate calculus, use elementary inference rules in a discrete math course, for example).
HYP 1: $P\vee \neg Q$
HYP 2: $Q \vee R$
HYP 3: $\neg R \vee S$
HYP 4: $P$
Thus $S$
Thanks, sorry for missing proper inference symbols. My professor is insistent that this works but I used contradiction and there was no contradiction.
I hope this can help:
p ∧(p ∨ ⊣q)⋀(q⋁r)⋀(⊣r⋁s)
≡p∧(q∨r)∧(⊣r⋁s)
Since p is always true, according to hypothesis we have:
((q∨r)∧⊣r)∨((q∨r)∧s)
≡((q∧⊣r)⋀F)⋁((q⋀s)⋁(r⋀s))
≡(q∧⊣r)⋁(q⋀s)⋁(r⋀s)
≡(q∧(⊣r⋁s))⋁(r⋀s)
As mentioned in the problem, ⊣r⋁s is always true. Now we have:
q∨(r∧s)≡T
(q∨r)∧(q∨s)≡T
And we know q∨r is always true, so:
q∨s≡T
This shows that we can conclude (q or s is true) and we can't conclude that s is true.
This is what I think about this problem, I hope I have solved it truly!!!