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Symbolic Logic A

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-- ShaughanLavine - 04 Nov 2004

-- BrendanMoloney? - 14 Dec 2004

Ok, here's my attempt at proving the first rule in exercise 4.5, pg 67. I don't know how to do substitution notation so i use / instead.---You use \frac{a}{x} (fraction) for a/x. I've fixed them below.-- ShaughanLavine - 15 Dec 2004

Suppose $\phi\models\psi$ . Then choose some arbitrary interpretation $\mathcal{I}$ * such that $\mathcal{I}\models\exists x\phi$ , we must show that $\mathcal{I}\models\exists x\psi$ . By the definition of satisfaction we know that there is an $a \in A$ such that $\mathcal{I}\frac{a}{x}\models\phi$ . Since $\phi\models\psi$ , $\mathcal{I}\frac{a}{x}\models\psi$ . Using the definition of satisfaction again, we get $\mathcal{I}\models\exists x\psi$ .

I'm not sure if the following part is really required, but i figure its better to be too explicit than not explicit enough. *It is not only not required, it is wrong: your reasoning is correct, but it presumes that $\phi$ is satisfiable. What needs to be shown is that if $\mathcal{I}\models\exists x\phi$ , then $\mathcal{I}\models\exists x\psi$ , which will certainly be true ("vacuously") if there is no $\mathcal{I}$ such that $\mathcal{I}\models\exists x\phi$ in the first place.* -- ShaughanLavine - 15 Dec 2004

*We know that such an interpretation exists because if $x\notin free(\phi)$ then using the Conincidence lemma we can conclude that for any $a \in A$ that $\mathcal{I}\frac{a}{x}\models\phi$ iff $\mathcal{I}\models\phi$ , if $x \in free(\phi)$ then we simply restrict ourselves to the interpretation $\mathcal{I}\frac{x}{x}$ hence $\mathcal{I}\frac{x}{x}\models\phi$ iff $\mathcal{I}\models\phi$ .

-- BrendanMoloney? - 14 Dec 2004 Hi there. I'm posting up my attempts at exercise 2.7 (page 63) for some feedback. Part a seems easy enough, but I think I'm doing something wrong in part b.

1.	$\vdash\Gamma,\phi_{1},\psi_{1}$	premise
2.	$\vdash\Gamma,\phi_{2},\psi_{2}$	premise
3.	$\vdash\Gamma,\phi_{1},\psi_{1}\lor\psi_{2}$	VS applied to 1
4.	$\vdash\Gamma,\phi_{2},\psi_{1}\lor\psi_{2}$	VS applied to 2
5.	$\vdash\Gamma,\phi_{1}\lor\psi_{1},\psi_{1}\lor\psi_{2}$	VA applied to 3 and 4

1.	$\vdash\Gamma,\phi_{1},\psi_{1}$	premise
2.	$\vdash\Gamma,\phi_{2},\psi_{2}$	premise
3.	$\vdash\Gamma,\phi_{1},\psi_{1}\land\psi_{2},\psi_{1}$	Ant. on 1
4.	$\vdash\Gamma,\phi_{1},\psi_{1}\land\psi_{2}$	Assm. on 3
5.	$\vdash\Gamma,\phi_{2},\psi_{1}\land\psi_{2},\psi_{2}$	Ant. on 2
6.	$\vdash\Gamma,\phi_{2},\psi_{1}\land\psi_{2}$	Assm. on 5
7.	$\vdash\Gamma,\phi_{1}\lor\phi_{2},\psi_{1}\land\psi_{2}$	VA on 4 and 6

-- JohnTrimble? - 14 Dec 2004

I don't believe that the above answer for 2.7b is correct. The problem is that the assumption of $\psi_{1}\land \psi_{2}$ was added to $\Gamma$ and then later it was concluded that somehow $\psi_{1} \land \psi_{2}$ was an assumption of $\Gamma$ ! I think the heart of the problem is that the Assumption rule was applied inappropriately. I'm fairly certain that the rule turns out to not be true, I'm currently working on a proof of this and will post it when I'm done (if I'm ever done). Quite right: line 4 is not a legitimate use of Assm. The rule doesn't depend on previous lines--"Assm on line 3" makes no sense. -- ShaughanLavine - 15 Dec 2004

-- JohnTrimble? - 14 Dec 2004

Below is my attempt to disprove 2.7b pg 63. There is a little bit of handwaving going on but I think that it is reasonable nonetheless (unless of course this rule turns out to be true):

Let $\Gamma{}=\lbrace\varphi_{1}\to\psi_{1},\varphi_{2}\to\psi_{2},\varphi_{1}\to\lnot\psi_{2},\varphi_{2}\to\lnot\psi_{1}\rbrace$
Notice that $\Gamma \varphi_{1}$ and $\Gamma \varphi_{2}$ are satisfiable.
Notice that $\Gamma\varphi_{1}\models\psi_{1}$ and $\Gamma \varphi_{2}\models\psi_{2}$
Let $\jmath$ be an abitrary interpretation such that $\jmath\models\Gamma\left(\varphi_{1}\lor\varphi_{2}\right)$
$\jmath\models\left(\varphi_{1}\lor\varphi_{2}\right)$
Either $\jmath\models\varphi_{1}$ or $\jmath \models \varphi_{2}$
Case 1:
$\jmath \models \varphi_{1}$
$\jmath\models\varphi_{1}\to\lnot\psi_{2}$
$\jmath \models \lnot \psi_{2}$
$\jmath \not \models \psi_{2}$
$\jmath\not\models\left(\psi_{1}\land\psi_{2}\right)$
Thus $\Gamma\left(\varphi_{1}\lor\varphi_{2}\right)\not\models\left(\psi_{1}\land\psi_{2}\right)$
Case 2: $\jmath\models\varphi_{2}$
Similar.

Therefore, the rule is not valid. $\Box$

-- BrendanMoloney? - 14 Dec 2004 I don't see what is different between what I did in my derivation, and what Shaughan does in the derivation at the bottom of the page. I use the Antecedent rule to add $\psi_{1}\land\psi_{2}$ to the antecedent and then using the Assumption rule to conclude $\psi_{1}\land\psi_{2}$ in the succedent. Shaughan uses the Antecedent rule to add $\neg\psi$ to the antecedent, and then uses the Assumption rule to conclude $\neg\psi$ in the succedent. This seems to be a common strategy in derivations, but it does seem a bit weird to me too.

-- BrendanMoloney? - 14 Dec 2004 On closer inspection I think my mistake was to not leave $\psi_{1}\land\psi_{2}$ in the antecedent after applying the Assumption rule. Then I would have to cancel it from the antecedent to get the desired result, which the above proof has convinced me is not possible.

Hi. This is not a solution, just a question about the term model. Ex 1.13 pg 79.

...Does $\mathcal{I}$ (the interpretation) depend on the inconsistent set $\Phi$ ? OK, so I know the answer is no, but I am confused as to why it is no. Would the interpretation not be different if $\Phi$ was in fact consistent? If yes, then doesn't $\mathcal{I}$ depend on the inconsistent set? If no, why not? Has anyone worked out this problem? If so please explain it to me. Thanks

For a given language there is a consistent theory $\Phi$ in the language such that $\mathcal{I}^\Phi$ is the same as $\mathcal{I}^\Psi$ for an inconsistent $\Psi$ in the language: Just let $\Phi$ be the set of all formulas satisfied by $\mathcal{I}^\Psi$ . The sense in which it is claimed that the interpretation does not depend on an inconsistent set of formulas is just that if Inc $\Phi$ and Inc $\Psi$ and $\Phi$ and $\Psi$ are sets of formulas of the same language, then $\mathcal{I}^{\Phi}=\mathcal{I}^\Psi$ .
-- ShaughanLavine - 09 Dec 2004

So, It's not asking if the interpretation depends on whether or not the set is consistent or inconsistent? It's just asking if it depends on which inconsistent set? This makes much, much, much more sense. Thanks

-- DesertMermaid? - 10 Dec 2004

P.S. Sorry I didn't use notation, I couldn't get the symbols on the same line with regular text and it just looked bad.
I fixed them. -- ShaughanLavine - 09 Dec 2004

Hello out there in Symbolic Logic World! I see nobody has posted anything here besides Shaughan. Well im going to change that big grin

post some problems I did involving substitution if anyone besides Shaughan reads this please feel free to comment on my work. Page 57 8.9 exercise (a)

$\exists{}x\exists{}y(Pxu\land{}Pyu)\frac{u}{x}\frac{u}{y}\frac{u}{v}=\exists{}x\exists{}y(Pxu\land{}Pyu)$

Solution:

$=\exists{}x\exists{}y(Pxu\frac{u}{y}\frac{u}{v}\land{}Pyu\frac{u}{x}\frac{u}{v)$

$=\exists x\exists y(Pxu\land{}Pyu)$

Here you can find other SubstitutionProblems? from 8.9 exercise

ShaughanLavine here. I've made one minor correction above: The original had $\equiv$ where it should have just had =. We're asserting in the metalanguage that certain metalinguistic expressions involving substitution notation represent the same formula of our official language. That is not an assertion in the language, but one made in English about it, and so the identity sign of English, namely =, is appropriate. The solution isn't quite correct. It should read as follows:

$=\exists{}x\exists{}y(Pxu\frac{u}{v}\land{}Pyu\frac{u}{v)$

$=\exists{}x\exists{}y(Pxu{}\land{}Pyu)$

Since both atomic formulas are in the scope of quantifications over both

and

, both of those substitutions get ``cancelled'' for both of them.

Here is an attempt at proving some rules on page 64 3.3 (b)
$\Gamma,\neg\phi,\psi$

$\Gamma,\neg\psi,\phi$

Solution
$\vdash\Gamma,\neg\phi,-,\psi$
$\vdash\Gamma,\neg\psi,\neg\phi,\psi$
$\vdash\Gamma,\neg\phi,\neg\phi,\neg\psi$
$\vdash\Gamma,\neg\psi,-,\phi$

I'am not sure how to add notes to the side after each line. Also those dashes should be blank spaces which I am too not able to produce. Yet again I don't know how to put things after the lines so here they are.
Line 1:premise
Line 2:(Ant) on line 1
Line 3:Assm
Line 4:(Ctr) on lines 2 and 3

ShaughanLavine again. Here is the derivation with the justifications next to the lines. I don't know why you think blanks or dashes were required, and so I've omitted them. The derivation is just like the one you gave. I've left both because it may be helpful to people trying to figure out how the Wiki works. The boxed $\neg\phi$ was a typo. I've replaced it with the correct $\neg\psi$ .

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

Here a few other JustificationsofRules? from 3.3 and derivations from exercise 3.6 page 65.

Latex rendering error!! dvi file was not created.

Topic revision: r1 - 23 Feb 2007 - 05:36:34 - ShaughanLavine

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