User:IssaRice/Logical induction notation

Term	Notation	Type	Definition	Notes
${\mathcal {F}}$-combination	$A$	${\mathcal {S}}\cup \{0,1\}\to {\mathcal {F}}_{n}$		Function application of an ${\mathcal {F}}$-combination uses square brackets instead of parentheses. Why? As far as I can tell, this is because each coefficient is in ${\mathcal {F}}$ so is itself a function. This means we have two senses of "application": we can pick out the specific coefficient we want (square brackets), or we can apply each coefficient to return something (parentheses).
Holdings from $T$ against ${\overline {\mathbb {P} }}$ (a $\mathbb {Q}$-combination)	$T({\overline {\mathbb {P} }})$	${\mathcal {S}}\cup \{0,1\}\to \mathbb {Q}$
Trading strategy	$T$	${\mathcal {S}}\cup \{1\}\to {\mathcal {E\!F}}$
Feature	$\alpha$	$[0,1]^{{\mathcal {S}}\times \mathbb {N} ^{+}}\to \mathbb {R}$ or equivalently $({\mathcal {S}}\times \mathbb {N} ^{+}\to [0,1])\to \mathbb {R}$ or equivalently ${\mathcal {F}}$

Example of a 5-strategy given on p. 18 of the paper:

$\underbrace {\left[(\neg \neg \phi )^{*5}-\phi ^{*5}\right]} _{\xi _{1}}\cdot (\phi -\phi ^{*5})+\underbrace {\left[\phi ^{*5}-(\neg \neg \phi )^{*5}\right]} _{\xi _{2}}\cdot \left(\neg \neg \phi -(\neg \neg \phi )^{*5}\right)$

Since the coefficients ($\xi _{1}$ and $\xi _{2}$) are in ${\mathcal {E\!F}}_{5}$, this is an ${\mathcal {E\!F}}_{5}$-combination. Let's call this 5-strategy $T_{5}$. We can pick out the coefficient for the $\phi$ term like $T_{5}[\phi ]=(\neg \neg \phi )^{*5}-\phi ^{*5}$. But since each coefficient is a feature (which is a function), we can also apply each coefficient to some valuation sequence ${\overline {\mathbb {V} }}$, like this:

$T_{5}({\overline {\mathbb {V} }})=\left[(\neg \neg \phi )^{*5}({\overline {\mathbb {V} }})-\phi ^{*5}({\overline {\mathbb {V} }})\right]\cdot (\phi -\phi ^{*5}({\overline {\mathbb {V} }}))+\left[\phi ^{*5}({\overline {\mathbb {V} }})-(\neg \neg \phi )^{*5}({\overline {\mathbb {V} }})\right]\cdot \left(\neg \neg \phi -(\neg \neg \phi )^{*5}({\overline {\mathbb {V} }})\right)$

Now each coefficient is a real number, so $T_{5}({\overline {\mathbb {V} }})$ is an $\mathbb {R}$-combination. Note that since $T_{5}\colon {\mathcal {S}}\cup \{1\}\to {\mathcal {E\!F}}_{5}$ is a function that takes a sentence or the number $1$ and ${\overline {\mathbb {V} }}$ is a valuation sequence (not a sentence or number), there appears to be a type error in writing $T_{5}({\overline {\mathbb {V} }})$. What is going on is that we aren't evaluating $T_{5}$ at ${\overline {\mathbb {V} }}$; rather, we are evaluating each coefficient of $T_{5}$, to convert the range of $T_{5}$ from ${\mathcal {E\!F}}_{5}$ to $\mathbb {R}$.

To summarize the types:

• $T_{5}\colon {\mathcal {S}}\cup \{1\}\to {\mathcal {E\!F}}_{5}$
• $T_{5}[\phi ]\in {\mathcal {E\!F}}_{5}$ in other words $T_{5}[\phi ]\colon [0,1]^{{\mathcal {S}}\times \mathbb {N} ^{+}}\to \mathbb {R}$
• $T_{5}({\overline {\mathbb {V} }})\colon {\mathcal {S}}\cup \{1\}\to \mathbb {R}$

If $T=c+\xi _{1}\phi _{1}+\cdots +\xi _{k}\phi _{k}\colon {\mathcal {S}}\cup \{1\}\to {\mathcal {E\!F}}_{n}$, then

$\mathbb {V} (T)=c+\xi _{1}\mathbb {V} (\phi _{1})+\cdots +\xi _{k}\mathbb {V} (\phi _{k})\in {\mathcal {E\!F}}_{n}$

and

$T({\overline {\mathbb {V} }})=c({\overline {\mathbb {V} }})+\xi _{1}({\overline {\mathbb {V} }})\phi _{1}+\cdots +\xi _{k}({\overline {\mathbb {V} }})\phi _{k}\colon {\mathcal {S}}\cup \{1\}\to \mathbb {R}$

and

$\mathbb {W} (T({\overline {\mathbb {V} }}))=c({\overline {\mathbb {V} }})+\xi _{1}({\overline {\mathbb {V} }})\mathbb {W} (\phi _{1})+\cdots +\xi _{k}({\overline {\mathbb {V} }})\mathbb {W} (\phi _{k})\in \mathbb {R}$

I think $\mathbb {W} (T({\overline {\mathbb {V} }}))=(\mathbb {W} (T))({\overline {\mathbb {V} }})$ but the former notation seems to be preferred in the paper.

The following is used in the Fixed Point Lemma (5.1.1):

Writing the $n$-strategy as

$T_{n}=\sum _{j=1}^{k}\xi _{j}\phi _{j}-\sum _{j=1}^{k}\xi _{j}\phi _{j}^{*n}$

we have

$\mathbb {V} (T_{n}(\mathbb {P} _{\leq n-1},\mathbb {V} ))=\sum _{j=1}^{k}\xi _{j}(\mathbb {P} _{\leq n-1},\mathbb {V} )\cdot \mathbb {V} (\phi _{j})-\sum _{j=1}^{k}\xi _{j}(\mathbb {P} _{\leq n-1},\mathbb {V} )\cdot \phi _{j}^{*n}(\mathbb {P} _{\leq n-1},\mathbb {V} )$

But $\phi _{j}^{*n}(\mathbb {P} _{\leq n-1},\mathbb {V} )=\mathbb {V} (\phi _{j})$ so the two sums cancel to obtain $0$.

See also

• https://machinelearning.subwiki.org/wiki/User:IssaRice/Logical_inductor_construction

External links

• https://arbital.com/p/logical_inductor_nota/
• https://arbital.com/p/logical_induction/