Prove Proof By Cases Using Laws of Inference

Question

I am a bit stuck proving the "proof by cases" using the basic laws of inference. Namely, I need to show that (P -> Q), ((not P) -> Q) implies Q.

How can I do that without making any leaps?

Answer 1

Hint

You have to use Excluded Middle : $P \lor \lnot P$ and then apply Proof by Cases:

$$ \cfrac{P \to Q \ \ \ \lnot P \to Q \ \ \ P \lor \lnot P}{\therefore Q} $$

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Another possibility is to use Material Implication to convert the premises into:

$\lnot P \lor Q$ and $P \lor Q$

followed by Resolution:

$$ \cfrac{\lnot P \lor Q \ \ \ \ P \lor Q}{Q \lor Q} $$

The conclusion follows by Idempotent laws.

Answer 2

Hint: Use the following rules of logic:

The generalized principle of proof by n-cases:

$$[P_1 \lor P_2 \lor P_3 \lor \cdots P_n]\land [[P_1\to Q] \land [P_2\to Q] \land \cdots [P_n \to Q]] \to Q$$

where the $P_i$ and $Q$ are any logical propositions.

The principle of the excluded middle:

$$P \lor \neg P$$

where $P$ is any logical proposition.