So I've been given one question from my colleague, which needs to prove using below premises in order to draw a certain conclusion (see hypothesis below).

I've been doing this from last night and by out of luck, I can't get the hypothesis right, even by referring to Wikipedia rules table here.

Premise 1: $(\neg P \rightarrow Q) \rightarrow R$

Premise 2: $P \rightarrow R$

Premise 3: $Q \rightarrow R$

Hypothesis: $Q \rightarrow (P \rightarrow R)$

From my observation, by using the first premise, I can draw out the second and third premises, but then I getting stuck there.

1) $(\neg P \rightarrow Q) \rightarrow R$

2) $\neg(\neg P \rightarrow Q) \vee R$

3) $\neg (\neg \neg P \vee Q) \vee R$

4) $\neg (P \vee Q) \vee R$

5) $(\neg P \wedge \neg Q) \vee R$

6) $(\neg P \vee R ) \wedge (\neg Q \vee R)$

7) $(P \rightarrow R ) \wedge (Q \rightarrow R)$

Can somebody point out what steps should I do in order to draw the hypothesis conclusion from above three premises?


From Premise 2: Suppose $(P\rightarrow R)$. Then whenever $Q$ is true, $(P\rightarrow R)$ is true (by our premise). So $Q\rightarrow (P\rightarrow R)$.

From Premise 3: Suppose $(Q\rightarrow R)$. Then whenever $Q$ is true and $P$ is true, $R$ is true (by our premise). So whenever $Q$ is true, $P\rightarrow R$ is true. So $Q\rightarrow (P\rightarrow R)$.

If you're not happy with this informal style of reasoning, here's another straightforward option: note that $Q\rightarrow (P\rightarrow R)$ is equivalent to $\lnot Q\lor (P\rightarrow R)$, which is equivalent to $\lnot Q\lor \lnot P\lor R$, so it follows immediately from $\lnot P\lor R$ (Premise 2) and from $\lnot Q\lor R$ (Premise 3) by the rule called Addition on Wikipedia's list (you're always free to add another term to a disjunction).

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I note that the Hypothesis can be inferred from each of the Premises alone ... Maybe that was the whole exercise?

OK, so here are 3 proofs, each starting with a different premise.

Proof from Premise 1:

  1. $(\neg P \rightarrow Q) \rightarrow R$ Premise 1

  2. $\quad Q$ Assumption

  3. $\quad \quad P$ Assumption

  4. $\quad \quad \quad \neg P$ Assumption

  5. $\quad \quad \quad Q$ Reiteration 2

  6. $\quad \quad \neg P \rightarrow Q$ $\rightarrow$ Intro 4-5

  7. $\quad \quad R$ $\rightarrow$ Elim 1,6

  8. $\quad P \rightarrow R$ $\rightarrow$ Intro 3-7

  9. $Q \rightarrow (P \rightarrow R)$ $\rightarrow$ Intro 2-8

Proof from Premise 2:

  1. $P \rightarrow R$ Premise 2

  2. $\quad Q$ Assumption

  3. $\quad P \rightarrow R$ Reiteration 1

  4. $Q \rightarrow (P \rightarrow R)$ $\rightarrow$ Intro 2-3

Proof from Premise 3:

  1. $Q \rightarrow R$ Premise 3

  2. $\quad Q$ Assumption

  3. $\quad \quad P$ Assumption

  4. $\quad \quad R$ $\rightarrow $ Elim 1,2

  5. $\quad P \rightarrow R$ $\rightarrow$ Intro 3-4

  6. $Q \rightarrow (P \rightarrow R)$ $\rightarrow$ Intro 2-5

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  • $\begingroup$ Can I know what is "Intro" and "Assumption"? I never heard or learn about these before in my logic class. $\endgroup$ – Mohd Shahril Apr 12 '17 at 5:24
  • $\begingroup$ @MohdShahril They are explained in the Wikipedia page you listed in your post. $\endgroup$ – Bram28 Apr 12 '17 at 12:22

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