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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.

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    Perhaps $P \wedge \neg Q$ was meant? Or maybe $\neg P \vee \neg Q$? – 2017-02-23
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    My professor is insistent that this works but I used contradiction and there was no contradiction :( – 2017-02-23
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    You're right; if we suppose $\neg S$, then we have $\neg R$ by hypothesis 3 and disjunction elimination, and thus $Q$ by hypothesis 2, ditto, but then disjunction elimination in hypothesis 1 gives us that $P$ has to be true, which is exactly what hypothesis 4 says. Maybe hypothesis 4 should have been $\neg P$? In any case, I'm almost certain there's a typo somewhere. – 2017-02-23
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    The conclusion does **not** follow; with a valuation $v$ such that $v(P)=v(Q)=$ **t** and $v(R)=v(S)=$ **f** all the hypotheses are satisfied and the conclusion is not. – 2017-02-23

1 Answers 1

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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!!!

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    It's not enough to analyze a single proof and conclude that you can't conclude something. You must provide a situation (truth assignment in this case) where the assumptions are satisfied by the conclusion is not satisfied. In fact, it gets worse; it is actually possible to show that some sentence is not provable over first-order logic (plus a few predicate symbols and function symbols) and yet there is a model for satisfying it, so unprovable does not even imply false. – 2017-02-25