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How to show $\vdash (((\psi \to \phi )\rightarrow \psi )\rightarrow \psi )$?Equivalently,$((\psi \to \phi )\rightarrow \psi )\vdash \psi $?Of course, by formal language $L$. The Axiom3 of $L$ is $(\neg \psi \to \neg \phi )\to (\phi \to \psi )$--not same to the answer already put on website. This question came from Logic for mathematicians written by A. G. Hammilton, in the chapter 2 exercise3.c.

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No big change needed...

In A.G. Hammilton, Logic for mathematicians's proof system we have:

Example 2.7 (page 31): $\vdash (\phi\to \phi)$.

Prop.2.8 (Deduction Theorem, page 32).

Prop.2.11 (page 35): (a) $\vdash (\lnot \phi \to (\phi \to \psi))$ and (b) $\vdash ((\lnot \phi\to \phi) \to \phi)$.

Now for the main proof :

  1. $(ϕ → ψ) → ϕ$ --- premise

  2. $\lnot \phi$ --- premise

  3. $\vdash \lnot \phi \to (\phi \to \psi)$ --- Prop.2.11 (a)

  4. $\phi \to \psi$ --- from 2) and 3) by MP

  5. $\phi$ --- from 1) and 4) by mp

  6. $(ϕ → ψ) → ϕ \vdash \lnot \phi \to \phi$ --- from 2) and 5) by Deduction Th

  7. $(ϕ → ψ) → ϕ \vdash ϕ$ --- Prop.2.11 (b) and MP

$\vdash ((ϕ → ψ ) → ϕ) → ϕ$ --- from t) by Deduction Th.

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  • $\begingroup$ Oh, I see. Thanks! $\endgroup$
    – YinFeng
    Commented Nov 9, 2020 at 10:30
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    $\begingroup$ @yinfeng - If you are satisfied with the answer, you can accept it. $\endgroup$ Commented Nov 9, 2020 at 10:34

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