Derivations

-- LostmyZ? - 05 Dec 2004 Here is an attempt at proving some rules on page 64 3.3 (c)
$\Gamma,\neg\phi,\neg\psi$
$\Gamma,\psi,\phi$

Solution

1. $\vdash\Gamma,\neg\phi,\hspace*{1em}\neg\psi$ premise
2. $\vdash\Gamma,\psi,\neg\phi,\neg\psi$ (Ant) on line 1
3. $\vdash\Gamma,\psi,\neg\phi,\psi$ Assm
4. $\vdash\Gamma,\psi,\hspace*{1em},\phi$ (Ctr) on lines 3 and 1

There was a typo in line 2 -- missing a negation. -- MarcJohansen? - 08 Dec 2004

Exercise 3.6 (a1)
$\Gamma,\phi$

$\Gamma,\neg\neg\phi$

Solution

1. $\vdash\Gamma,\hspace*{1em},\phi$ premise
2. $\vdash\Gamma,\neg\phi,\phi$ (Ant) on line 1
3. $\vdash\Gamma,\neg\phi,\neg\psi$ Assm
4. $\vdash\Gamma,\neg\neg\phi,\neg\psi$ (Ant) on lines 3
5. $\vdash\Gamma,\neg\neg\phi,\neg\neg\psi$ Assm
6. $\vdash\Gamma,-,\neg\neg\phi$ (Ctr') on lines 4 and 5

Lines 3 & 5 above aren't valid instances of the assumption rule. How's this?

1 $\Gamma, \phi $ premise
2 $\Gamma, \neg\phi, \phi $ ant (line 1)
3 $\Gamma, \neg\phi, \neg\phi $ assm
4 $\Gamma, \neg\phi, \neg\neg\phi $ ctr' (lines 2 & 3)
5 $\Gamma, \neg\neg\phi, \neg\neg\phi $ assm
6 $\Gamma, \neg\neg\phi $ pc (lines 4 & 5)

-- MarcJohansen? - 08 Dec 2004

3.6 (b)

I'll use "+" as shorthand for the equivalences from p. 35.

1 $\Gamma, \phi $ premise
2 $\Gamma, \psi $ premise
3 $\Gamma, \neg\phi\lor\neg\psi, \phi $ ant (line 1)
4 $\Gamma, \neg\phi\lor\neg\psi, \neg\neg\phi $ 3.6 a1 (line 3)
5 $\Gamma, \neg\phi\lor\neg\psi, \neg\phi\lor\neg\psi $ assm
6 $\Gamma, \neg\phi\lor\neg\psi, \neg\psi $ 3.4 (lines 4 & 5)
7 $\Gamma, \neg\phi\lor\neg\psi, \psi $ ant (line 2)
8 $\Gamma, \neg(\neg\phi\lor\neg\psi) $ ctr (lines 6 & 7)
9 $\Gamma, \phi\land\psi $ + (line 8)

-- MarcJohansen? - 08 Dec 2004

3.6 (c)

1 $\Gamma, \phi, \psi $ premise
2 $\Gamma, \phi, \neg\phi\lor\psi $ $\lor$S (line 1)
3 $\Gamma, \neg\phi, \neg\phi $ assm
4 $\Gamma, \neg\phi, \neg\phi\lor\psi $ $\lor$S (line 3)
5 $\Gamma, \neg\phi\lor\psi $ pc (lines 2 & 4)
6 $\Gamma, \phi\rightarrow\psi $ + (line 5)

-- MarcJohansen? - 08 Dec 2004

5.5. (b2)

1 $\Gamma, \phi\frac{y}{x} $ premise
2 $\Gamma, \neg\phi\frac{y}{x}, \neg\phi\frac{y}{x} $ assm
3 $\Gamma, \exists{x}\neg\phi, \neg\phi\frac{y}{x} $ if y is not free in the above sequent $\exists$ A (line 2)
4 $\Gamma, \phi\frac{y}{x}, \neg\exists{x}\neg\phi $ cp (d) (line 3)
5 $\Gamma, \neg\exists{x}\neg\phi $ Ch (lines 1 & 4)
6 $\Gamma, \forall{x}\phi $ + (line 5)

-- MarcJohansen? - 08 Dec 2004

Proofs of Correctness

Number 1
$\Gamma,(\phi \land \psi)$

$\Gamma,\phi$

Solution
Suppose $\Gamma\models(\phi \land \psi)$. We must show that $\Gamma\models\phi$. Let $\mathcal{I}$ be any interpretation such that $\mathcal{I}\models\Gamma$. So $\mathcal{I}\models(\phi \land \psi)$. If $\mathcal{I}\models(\phi \land \psi)$ then $\mathcal{I}\models\phi$ and $\mathcal{I}\models\psi$ by the definition of satisfaction. If $\mathcal{I}\models(\phi \land \psi)$ then, since $\Gamma\models(\phi \land \psi)$ it follows that Since $\mathcal{I}\models\phi$, and $\mathcal{I}$ was an arbitrary interpretation that satisfies $\Gamma$, and so $\Gamma\models\phi$.
Forgive me I don't know the proper text to get the script I.
You use \mathcal{I}. I fixed them. I also corrected this problem. The others, like this one, are basically correct, but they contain similar mistakes. I'll leave them for you to fix.
-- ShaughanLavine - 09 Dec 2004

Number 2
$\Gamma,\phi$ $\Gamma,\psi$
$\Gamma,(\phi \land \psi)$

Solution
Suppose $\Gamma\models\phi$ and $\Gamma\models\psi$. We must show that $\Gamma\models(\phi \land \psi)$. Let $\mathcal{I}$ be any interpretation that $\mathcal{I}\models\Gamma$. So $\mathcal{I}\models\phi$ and $\mathcal{I}\models\psi$. If $\mathcal{I}\models\phi$ then, since $\Gamma\models\phi$ and $\Gamma\models\psi$ it follows that $\mathcal{I}\models(\phi \land \psi)$ so $\Gamma\models(\phi \land \psi)$.

Number 3
$\Gamma,\neg\neg\phi$
$\Gamma,\phi$

Solution
Suppose $\Gamma\models\neg\neg\phi$. We must show that $\Gamma\models\phi$. Let $\mathcal{I}$ be any interpretation that$\mathcal{I}\models\Gamma$. So $\mathcal{I}\models\neg\neg\phi$. If $\mathcal{I}\models\neg\neg\phi$ then, since $\Gamma\models\neg\neg\phi$ by the negation clause if $\mathcal{I}\models\neg\neg\phi$ then $\mathcal{I}\not\models\neg\phi$ and again by the negation clause if$\mathcal{I}\not\models\neg\phi$ then $\mathcal{I}\models\phi$. It follows that $\mathcal{I}\models\phi$ so $\Gamma\models\phi$