Friday, October 28, 2016
Sunday, October 23, 2016
some notes from Hamilton Logic.
The formal system L of statement calculus is defined by the following:
1. Alphabet of symbols(infinite):
$$
\neg ,(,),p_{ 1 },p_{ 2 },p_{ 3 },\quad ...
$$
2. Set of wfs. Instead of specifying the set explicitly we give an inductive rule in three parts.
(i) $p_{i}$ is a wf. for each $i>1$.
(ii) if $\phi$ and $\chi$ are wfs. then $(\neg\phi)$ and $(\phi \rightarrow \chi)$ are wfs.
(ii) The set of all wfs. is generated by (i) and (ii).
3. Axioms.There are infinitely many axioms, so we cannot list them all. However we can specify all of them by means of three axiom schemes.For any wfs.
$\phi$,$\chi$, $\zeta$, the following wfs. are axioms of L:
(L1) $(\phi \rightarrow (\chi \rightarrow \phi))$
(L2) $(\phi \rightarrow (\chi \rightarrow \zeta)) \rightarrow ((\phi \rightarrow \chi) \rightarrow (\phi \rightarrow \zeta))$
1. Alphabet of symbols(infinite):
$$
\neg ,(,),p_{ 1 },p_{ 2 },p_{ 3 },\quad ...
$$
2. Set of wfs. Instead of specifying the set explicitly we give an inductive rule in three parts.
(i) $p_{i}$ is a wf. for each $i>1$.
(ii) if $\phi$ and $\chi$ are wfs. then $(\neg\phi)$ and $(\phi \rightarrow \chi)$ are wfs.
(ii) The set of all wfs. is generated by (i) and (ii).
3. Axioms.There are infinitely many axioms, so we cannot list them all. However we can specify all of them by means of three axiom schemes.For any wfs.
$\phi$,$\chi$, $\zeta$, the following wfs. are axioms of L:
(L1) $(\phi \rightarrow (\chi \rightarrow \phi))$
(L2) $(\phi \rightarrow (\chi \rightarrow \zeta)) \rightarrow ((\phi \rightarrow \chi) \rightarrow (\phi \rightarrow \zeta))$
(L3) $((\neg \phi) \rightarrow (\neg \chi)) \rightarrow (\chi \rightarrow \phi)$
Note that each axiom scheme has infinitely many 'instances', as $\phi,\chi,\zeta$rane over all wfs of L.
4. Rules of deduction. In L there is only one rule of deduction. namely modus ponens(abbreviated by MP). It says: From $\phi$ and $(\phi \rightarrow \chi)$, $\chi$ is a direct consequence, where $\phi$ and $\chi$ are wfs. of L.
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Definition 2.2
A proof in L is a sequence of wfs. $\phi_{1}, ..., \phi_{n}$ such for each i $(1\leq i\leq n)$, either $\phi_{i}$ is anaxiom of L or $\phi_{i}$ follows from two previus members of the sequence, say $\phi_{j}$ and $\phi_{k}$ $(j<i,k<i)$ as a direct consequence using the rule of deduction MP. Such a proof will be referred to as a proof of $\phi_{n}$ in L. and $\phi_{n}$ is said to be a theorem of L.
Remark 2.3
(a) in the above deifintion, observe that $\phi_{i}$ and $\phi_{k}$ must necessarily be of the form $\chi$ and $\chi \rightarrow \phi_{i}$, or vice versa.
(b) If $\phi_{1}, ..., \phi_{n}$ is a proof in L and $k<n$ then $\phi_{1}, ..., \phi_{k}$ is also a proof in L(its clearly satisfies the definition), and so $\phi_{k}$ is a theorem of L.
(c) Axioms of L are certainly theorem of L. Their proofs in L are one-member sequences.
----------------------------------------------------------------------------------
Definiton 2.5
Let $\Gamma $ be a set of wfs. of L (which may or may not be axioms or theorems of L). A sequence $\phi_{1},...,\phi_{n}$ of wfs. of L is a deduction from $\Gamma$ if for each i $(1\leq i \leq n)$
, one of the following holds:
(a) $\phi_{i}$ is an axiom of L,
(b) $\phi_{i}$ is a member of $\Gamma$, or
(c) $\phi_{i}$ follows from two previous members of the sequence as direct consequence using MP.
So a deduction from $\Gamma$ is just a 'proof' in which the members of $\Gamma$ are considered as temporary axioms.
the last Member $\phi_{n}$, of a sequence which is a deduction from $\Gamma$ is said to be deducible from $\Gamma$, or to be a consequence of $\Gamma$ in L.
if a wf. $\phi$ is the last member of some deduction from $\Gamma$, we say $\Gamma$ yields $\Gamma\vdash_{L} \phi$.
Notice that a theorem of L is deducible from empty set of wfs.(a proof in L is a deduction from $\emptyset$), so if $\phi$ is a theorem of Lwe may write $\emptyset \vdash_{L} \phi $,or more normally for brevity, just $\vdash_{L} \phi $
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Proposition 2.8(The Deduction Theorem)
if $\Gamma \cup \{ \phi \} \vdash \chi$ then $\Gamma \vdash_{l} (\phi \rightarrow \chi $, where $\phi$ and $\chi$ are wfs. of L, and $\Gamma$ is a set of wfs. of L(possibly empty).
Proof. The proof is by induction on the number of wfs, in seqence foming the deduction...
Note that each axiom scheme has infinitely many 'instances', as $\phi,\chi,\zeta$rane over all wfs of L.
4. Rules of deduction. In L there is only one rule of deduction. namely modus ponens(abbreviated by MP). It says: From $\phi$ and $(\phi \rightarrow \chi)$, $\chi$ is a direct consequence, where $\phi$ and $\chi$ are wfs. of L.
-----------------------------------------------------------------------------
Definition 2.2
A proof in L is a sequence of wfs. $\phi_{1}, ..., \phi_{n}$ such for each i $(1\leq i\leq n)$, either $\phi_{i}$ is anaxiom of L or $\phi_{i}$ follows from two previus members of the sequence, say $\phi_{j}$ and $\phi_{k}$ $(j<i,k<i)$ as a direct consequence using the rule of deduction MP. Such a proof will be referred to as a proof of $\phi_{n}$ in L. and $\phi_{n}$ is said to be a theorem of L.
Remark 2.3
(a) in the above deifintion, observe that $\phi_{i}$ and $\phi_{k}$ must necessarily be of the form $\chi$ and $\chi \rightarrow \phi_{i}$, or vice versa.
(b) If $\phi_{1}, ..., \phi_{n}$ is a proof in L and $k<n$ then $\phi_{1}, ..., \phi_{k}$ is also a proof in L(its clearly satisfies the definition), and so $\phi_{k}$ is a theorem of L.
(c) Axioms of L are certainly theorem of L. Their proofs in L are one-member sequences.
----------------------------------------------------------------------------------
Definiton 2.5
Let $\Gamma $ be a set of wfs. of L (which may or may not be axioms or theorems of L). A sequence $\phi_{1},...,\phi_{n}$ of wfs. of L is a deduction from $\Gamma$ if for each i $(1\leq i \leq n)$
, one of the following holds:
(a) $\phi_{i}$ is an axiom of L,
(b) $\phi_{i}$ is a member of $\Gamma$, or
(c) $\phi_{i}$ follows from two previous members of the sequence as direct consequence using MP.
So a deduction from $\Gamma$ is just a 'proof' in which the members of $\Gamma$ are considered as temporary axioms.
the last Member $\phi_{n}$, of a sequence which is a deduction from $\Gamma$ is said to be deducible from $\Gamma$, or to be a consequence of $\Gamma$ in L.
if a wf. $\phi$ is the last member of some deduction from $\Gamma$, we say $\Gamma$ yields $\Gamma\vdash_{L} \phi$.
Notice that a theorem of L is deducible from empty set of wfs.(a proof in L is a deduction from $\emptyset$), so if $\phi$ is a theorem of Lwe may write $\emptyset \vdash_{L} \phi $,or more normally for brevity, just $\vdash_{L} \phi $
---------------------------------------------------------------------------------------
Proposition 2.8(The Deduction Theorem)
if $\Gamma \cup \{ \phi \} \vdash \chi$ then $\Gamma \vdash_{l} (\phi \rightarrow \chi $, where $\phi$ and $\chi$ are wfs. of L, and $\Gamma$ is a set of wfs. of L(possibly empty).
Proof. The proof is by induction on the number of wfs, in seqence foming the deduction...
There is a number which it's natural logarithm is equal to 1
Prove that there is a number $e$ at which the area under the curve of $\frac{1}{x}$ is equal to 1, and show that the number $e$ should be $2<e<3$;
It is obvious from the graph picture that we can have an area of the size of 1;
first of all we check x=2,we can draw a rectangle from x=1 to x=2, because the equation is 1/x, at x=1 the equation gives us y=1=1/1 which is the high of our rectangle(in this case it would be square). the area of our shape(gray region) would be 1*1=1; and we can see that the actual area of the curve at this region is less than our square(because of concavity); therefore at x=2 we can not have the area 1 and x must be more than 1;
now we have to prove that the x can not be more than 3. if we draw tangent line of equation 1/x at x=2, and then calculate the area under our tangent line equation we can judge about e.
the line equation at x=2 would be :
$$
f(x)=\frac { 1 }{ x } ,\quad f(x)'=-\frac { 1 }{ x^{ 2 } } ,\quad f(2)'=-\frac { 1 }{ 4 } \\ \\ l(x)=f(2)+f(2)'(x-2)\\ l(x)=\frac { 1 }{ 2 } -(\frac { 1 }{ 4 } )(x-2)
$$
the area of the l(x) from 1 to 3 gives us the trapezoid, we can calculate the area in two way, using classic geometry or definite integral, we do it by the second method:
$$
\\ \int _{ 1 }^{ 3 }{ l(x)dx } =\int _{ 1 }^{ 3 }{ [\frac { 1 }{ 2 } -(\frac { 1 }{ 4 } )(x-2)]dx } =[x-\frac { 1 }{ 8 } x^{ 2 }]\overset { 3 }{ \underset { 1 }{ \quad } } =1\\
$$
the area of the trapezoid is equal to 1, but we can see from the picture that the area of actual curve in this interval is more than the trapezoid area. the exact area is $ trapezoid area(=1) + \Delta s$. at the points greater than 3 the area must be more than 1; therefore $e$ should be less than 3;
Notice that the number e is very useful, we know some important properties of natural logarithm like: $ln(a^{x})=x.ln(a)$ (we can prove it easily by the fact that the functions $ln(ax)$ and $ln(x)$ have the same derivative), now we have a substitute for b at which the ln is equal to 1, namely $ln(e)=1$. therefore we achieved something wonderful! the inverse function of $ln(x)$. remember that two inverse function have this property : $f(f(x))^{-1}=x$. now we have this property!
$ln(e^{x}) =x.ln(e)=x$ , therefore the functions $e^{x}$ and $ln(x)$ are inverses of each other.
It is obvious from the graph picture that we can have an area of the size of 1;
first of all we check x=2,we can draw a rectangle from x=1 to x=2, because the equation is 1/x, at x=1 the equation gives us y=1=1/1 which is the high of our rectangle(in this case it would be square). the area of our shape(gray region) would be 1*1=1; and we can see that the actual area of the curve at this region is less than our square(because of concavity); therefore at x=2 we can not have the area 1 and x must be more than 1;
now we have to prove that the x can not be more than 3. if we draw tangent line of equation 1/x at x=2, and then calculate the area under our tangent line equation we can judge about e.
the line equation at x=2 would be :
$$
f(x)=\frac { 1 }{ x } ,\quad f(x)'=-\frac { 1 }{ x^{ 2 } } ,\quad f(2)'=-\frac { 1 }{ 4 } \\ \\ l(x)=f(2)+f(2)'(x-2)\\ l(x)=\frac { 1 }{ 2 } -(\frac { 1 }{ 4 } )(x-2)
$$
the area of the l(x) from 1 to 3 gives us the trapezoid, we can calculate the area in two way, using classic geometry or definite integral, we do it by the second method:
$$
\\ \int _{ 1 }^{ 3 }{ l(x)dx } =\int _{ 1 }^{ 3 }{ [\frac { 1 }{ 2 } -(\frac { 1 }{ 4 } )(x-2)]dx } =[x-\frac { 1 }{ 8 } x^{ 2 }]\overset { 3 }{ \underset { 1 }{ \quad } } =1\\
$$
the area of the trapezoid is equal to 1, but we can see from the picture that the area of actual curve in this interval is more than the trapezoid area. the exact area is $ trapezoid area(=1) + \Delta s$. at the points greater than 3 the area must be more than 1; therefore $e$ should be less than 3;
Notice that the number e is very useful, we know some important properties of natural logarithm like: $ln(a^{x})=x.ln(a)$ (we can prove it easily by the fact that the functions $ln(ax)$ and $ln(x)$ have the same derivative), now we have a substitute for b at which the ln is equal to 1, namely $ln(e)=1$. therefore we achieved something wonderful! the inverse function of $ln(x)$. remember that two inverse function have this property : $f(f(x))^{-1}=x$. now we have this property!
$ln(e^{x}) =x.ln(e)=x$ , therefore the functions $e^{x}$ and $ln(x)$ are inverses of each other.
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