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I'm trying to solve this exercise.

If $\phi, \psi$ are propositional formulas, prove $\vdash (((\phi \to \psi) \to \phi) \to \phi)$.

I was wondering if there was a "simple" proof to this (I'm aware of the proof Using the Fitch system, how do I prove $((p \implies q) \implies p)\implies p$?), and most importantly, how to motivate a proof to this exercise. (For me, the motivation is the most important bit! I already have a proof)

I asked such a question before, with a much simpler exercise, and the answer was "try it out in words first, and then write it symbolically!" which worked very well until coming across this example. Currently I have a proof to this, which I brute forced; to me it feels extremely mechanical and non-replicable but for luck. I'm not sure if this is relevant to the question, but I'll attach it as a picture. deductionproof

Edit: I actually don't know what the expression "proof system" means. Sorry; I was only told that this was "natural deduction", and the rules for $\land, \lnot, \rightarrow , \leftrightarrow, \perp$, etc were given.

Albert
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3 Answers3

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Given the last instance of the material conditional then the last rule to use must be an instance of the deduction theorem (or ($\to I$)), so for the solution a good strategy is to assume the antecedent: $(\phi \to \psi)\to \phi$, turn it into its equivalent disjunctive version: $\lnot(\phi \to \psi) \lor \phi$; then assume the two disjuncts to derive the formula $\phi$ on both sides and dischard the two assumptions $(\lor E)$. then you can make use of $(\to I)$, dischard the original assumption and derive the desired sequent.

Simone
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If $\phi$ is false, then $\phi \to \psi$ is true, but then since $(\phi \to \psi) \to \phi,$ from modus ponens $\phi$ would be true. So that much shows $\phi$ cannot be false, i.e. $\phi$ must be true.

This isn't a formal proof, but could be turned into one in some systems.

coffeemath
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Your proof has a few redundant steps. Let us rebuild it.

You have the miniproof: $$\dfrac{\dfrac{\dfrac{\lnot\phi\quad[\phi]^3}{\bot}{\small\lnot\mathsf E}}{\psi}{\small\bot}}{\phi\to\psi}{\small{\to}\mathsf I^3}$$

And you know that you want to end with a conditional introduction:

$$\dfrac{\dfrac{[(\phi\to\psi)\to\phi]^1\\\qquad\quad\vdots}{\phi}{}}{((\phi\to\psi)\to\phi)\to\phi}{\small{\to}\mathsf I^1}$$

Well simply combining them to use a conditional elimination will derive $\phi$ but ...

$$\dfrac{\lower{1.5ex}{(\phi\to\psi)\to\phi}\quad\dfrac{\dfrac{\dfrac{\overset{\color{red}\downarrow}{\lnot\phi}\quad[\phi]^3}{\bot}{\small\lnot\mathsf E}}{\psi}{\small\bot}}{\phi\to\psi}{\small{\to}\mathsf I^3}}{\phi}{\small{\to}\mathsf E}$$

Uh oh! You cannot yet introduce the conditional unless you can discharge that undischarged $\lnot\phi$ and still derive $\phi$.

Fortunately, if you could derive a contradiction you may then use RAA to do that.

Okay, can you derive that contradiction?

Uh, yes, obviously: the assumption $\lnot\phi$ and sub-conclusion $\phi$ do contradict.

So...

$$\dfrac{\dfrac{\dfrac{\dfrac{\lower{1.5ex}{[(\phi\to\psi)\to\phi]^1}\quad\dfrac{\dfrac{\dfrac{[\lnot\phi]^2\quad[\phi]^3}{\bot}{\small\lnot\mathsf E}}{\psi}{\small\bot}}{\phi\to\psi}{\small{\to}\mathsf I^3}}{\phi}{\small{\to}\mathsf E}}{\bot}{\small\lnot\mathsf E}}{\phi}{\small\textsf{RAA}^2}}{((\phi\to\psi)\to\phi)\to\phi}{\small{\to}\mathsf I^1}$$

Graham Kemp
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