I am reading "Intermediate Logic" by David Bostock and I'm having some issues with the excercises. The principles of "cutting", "thinning" and "assumption" have been introduced as well as entailment principles such as the implication one stated below:

Implication principle

$\Gamma \models \psi \rightarrow\phi $ iff $\Gamma, \psi \models \phi$

The excercise 2.5.4 states:

Show that the principle of implication and negation implies: (and vice-versa)

$(1)\ \psi\models \phi \rightarrow \psi $ and $\lnot\phi\models\phi\rightarrow\psi$

$(2)\ \phi,\phi\rightarrow\psi\models \psi $

I'm not sure how to parse this excercise but I figure I'm supposed to show that (2) follows from (1) given the principles stated.

What I've tried is, starting off with (1):

$(1)\leftrightarrow\psi,\phi\models\psi \ \leftrightarrow \ $ ("cutting-rule")$\ \leftrightarrow\ \phi \rightarrow \psi,\phi\models\psi\equiv (2)$.

But I don't think this is quite correct.



For (2), consider $\Gamma = \{ \phi \to \psi \}$; thus, the Conditional principle amounts to :

$\phi \to \psi \vDash \phi \to \psi \ \text{ iff } \ \phi \to \psi, \phi \vDash \psi$.

The left-hand side is licensed by the Assumptions principle.

For (1a) : $ψ⊨ϕ→ψ$, Assumptions, Thinning and Conditional are needed :

(i) $ψ⊨ψ$

(ii) $ϕ, ψ⊨ψ$

(iii) $ψ⊨ϕ→ψ$.

  • $\begingroup$ So the correct way to cypher these excercises is to construct the statements using only the basic three principles in conjunction with the allowed secondary principles (implications, conjunction etc.)? $\endgroup$ – Strange Brew Oct 27 '15 at 12:11
  • 1
    $\begingroup$ @StrangeBrew - yes; see page 34, how some "derived" principles, like ex falso quodlibet are proved from the "basic" ones. The ex falso q principle is needed (as suggested in Ex.2.5.4.) to prove (2b). $\endgroup$ – Mauro ALLEGRANZA Oct 27 '15 at 12:14
  • $\begingroup$ Okay, I see. So to prove (1b) I could state (i)$\phi \models\phi $ (ii)$\lnot \phi, \phi\models$ (iii)$\lnot \phi, \phi\models \psi$ (iv)$\lnot \phi \models\phi \rightarrow \psi$? $\endgroup$ – Strange Brew Oct 27 '15 at 12:17
  • $\begingroup$ @StrangeBrew - exactly. $\endgroup$ – Mauro ALLEGRANZA Oct 27 '15 at 13:39

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