I already know how to prove $((p ⇒ q) ⇒ (p ⇒ r))$, given $((p ⇒ q) ⇒ r)$. I'm having some difficulties with the proof in the title, however. Help would be appreciated
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OK, what do you think the basic proof strategy should be, given your goal? – Bram28 Jan 27 '17 at 18:39
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To make this proof, you only need to use "new premise", "$\implies$ introduction" , and "$\implies$ elimination". – DanielV Jan 27 '17 at 21:06
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Using Fitch System (Let the letters inside brackets, ie [a], represent the levels of assumptions):
- $(p \implies (q \implies r))$ Premise
- $(p \implies q)$ Assumption [a]
- $p$ Assumption[b]
- $q \implies r$ Implication Elimination 1,3 [b]
- $q$ Implication Elimination 2,3 [b]
- $r$ Implication Elimination 4,5 [b]
- $p \implies r$ Implication Introduction 3,6 [a]
- $(p \implies q) \implies (p \implies r)$ Implication Introduction 2,7
$\therefore{(p \implies (q \implies r)) \implies ((p \implies q) \implies (p \implies r))}$
The trick for this one is that you want to start with an assumptions that will enable you to, after an Implication Introduction, reach your goal. In this case we know that if we make an assumption $(p \implies q)$, then when we do Implication Introduction we will get something that looks like $(p \implies q) \implies ?$. The next step would be to somehow form $(p \implies r)$ with another nested assumption.
Here is a good link to help you understanding better how to prove with Fitch. See Chapter 4.
Kunashu
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I'll write in a style quite similar to Fitch's.
If $P \to ( Q \to R )$:
If $P \to Q$:
If $P$:
...
$R$.
$P \to R$.
$( P \to Q ) \to ( Q \to R )$.
I'll leave you to fill in the blanks. The structure I've given above falls naturally out of the goal. All that is left is to use the assumptions in the inner context to get from $P$ to $R$. That should not be hard.
