Symbolic Logic Proof -- Typo?

Question

I was working through some proofs in my symbolic logic book when I came across the following problem:

$$\text{Given:}$$ $$\\ (A \lor B) \implies(C \lor D)$$ $$\\ (C \implies W) \land (D \implies \neg A)$$ $$\\A \land \neg W$$ $$\text{Prove:}$$ $$\neg(C \lor D)$$

My Proof:

By Simplification $$A$$

By Addition $$A \lor B$$

By Modus Ponens $$C \lor D$$

This, however, is the negation of the expression I was asked to prove and (if my reasoning is correct) $\neg(C \lor D)$ cannot be inferred from the premises. Was there a mistake in the book or in my reasoning? Thanks.

Answer 1

Yes, I suspect there's a typo there - those hypotheses are inconsistent.

Although that said, technically the problem does work as stated - an inconsistent theory proves everything. So it is possible that this was intensional.

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Here's how the inconsistency comes about roughly:

• Since by $(3)$ we know $A$ is true, $A\vee B$ is true and so $(1$) tells us that either $C$ or $D$ is true.

• $(2)$ tells us that if $C$ were true, then $W$ would be true; since $(3)$ tells us that $W$ is false, we know that $C$ is false.

• ... And so $D$ must be true. But then $(2)$ tells us that $\neg A$ is true, and this contradicts $(3)$.

Now as noted above, since an inconsistent system proves everything this is technically a valid exercise; but I don't really think this was what was intended.

Answer 2

Your proof that $C\lor D$ follows is correct. However, $\lnot(C\lor D)$ also follows, so the premises are inconsistent.

We can prove it by contradiction. (I will be very informal so as to allow you to construct the formal proof.)

If $C$ holds, then since we have $C\to W,$ $W$ holds, but we also have $\lnot W.$ If $D$ holds, then since we have $D\to \lnot A,$ $\lnot A$ holds. But we also have $A.$