Symbolic Logic:Programming:Value Set Proof

Set Membership Equivalence

 * $$x \in S \iff V(x) = v(\{ z\in S : z :: \{x = z\} \}) \! $$

Proof
Firstly $$v(\{ z\in S : z :: \{x = z\} \}) \! $$ is a Value Set because,


 * A1 is true


 * A3
 * $$W(v(\{ z\in S : z :: \{x = z\} \})) = \{z\in S : \{x = z\}\} \! $$
 * $$(\bigcup_{z\in S} : \{x = z\}) = \{x \in S\}= \{true\} \! $$


 * A4
 * $$\forall z \in {V(x)}_t : z_p \subset \{ x = z_v \} \! $$

is,
 * $$\forall z \in \{ z\in S : z :: \{x = z\} \} : z_p \subset \{ x = z_v \} \! $$

simplifies to,
 * $$\forall z \in S : \{x = z\} \subset \{ x = z \} \! $$

which is true,

Secondly using A2,
 * $$S(v(\{ z\in S : z :: \{x = z\} \})) = S \! $$

so every Value Set Y with S(Y) = Y must equal $$v(\{ z\in S : z :: \{x = z\} \})\!$$

Functions of a Value Set
The following theorem gives the function on a value set of a type using the function of the type.


 * (B1) $$f(X) = v(\{y\in X_t: f(y_v)::y_p\})\!$$

Proof
Let,
 * $$K = v(\{y\in X_t: f(y_v)::y_p\}) \} \!$$

firstly we need to prove that K satisfies the axioms of a Value Set, if X is a value set,

Step 1
To prove that axiom A3 holds,
 * $$W(K) = W(v(\{y\in X_t: f(y_v)::y_p\})) = \{y\in X_t: y_p\} = W(X) \!$$

so,
 * $$ \bigcup_{w \in W(K)} : w = \bigcup_{w \in W(X)} : w = \alpha \! $$

Step 2
From (A4) applied to X,
 * $$\forall z \in X : z_p \subset \{ x = z_v \} \! $$

which implies,
 * $$\forall z \in X : z_p \subset \{ f(x) = f(z_v) \} \! $$

Which is (A4) applied to K.

Step 3
As $$K\!$$ obeys the definition of a Value Set now prove that it equals $$f(X)\!$$. Let $$X = V(x)\!$$.

From A6,
 * $$V(f(x)) = f(V(x)) \!$$

From A1,
 * $$x \in S(V(x))\!$$

which implies,
 * $$f(x) \in \{y \in S(V(x)):f(y)\} \!$$

so,
 * $$S(V(f(x)) = \{y \in S(V(x)):f(y)\} \!$$

but also by simplifying,
 * $$S(K) = S(v(\{y\in X_t: f(y_v)::y_p\})) = \{y \in S(X): f(y)\} \!$$

so from A2,
 * $$S(V(f(x)) = S(K) \implies V(f(x)) = K \!$$

gives,
 * $$V(f(x)) = f(V(x)) = f(X) = K \!$$

Function of Multiple Parameters
If $$F$$ is a value set of functions on a value set of a values,


 * (B2) $$F(X) = v(\{f\in F_t, x\in X_t: f_v(x_v)::f_p\cap x_p\}) \!$$

Let,
 * $$K = v(\{f\in F_t, x\in X_t: f_v(x_v)::f_p\cap x_p\}) \} \!$$

firstly we need to prove that K satisfies the axioms of a Value Set, if X is a value set,

Step 1
To prove that axiom A3 holds,
 * $$W(K) = W(v(\{f\in F_t, x\in X_t: f_v(x_v)::f_p\cap x_p\})) = \{f\in F_t, x\in X_t: f_p\cap x_p\} \!$$

so,
 * $$ (\bigcup_{w \in W(K)} : w) = (\bigcup_{f\in F_t, x\in X_t}: f_p\cap x_p) \! $$

gives,
 * $$ \bigcup_{w \in W(K)} : w = (\bigcup_{f\in F_t}: f_p)\cap(\bigcup_{x\in X_t}: f_p\cap x_p) = \alpha \cap \alpha = \alpha \! $$

which is A3 applied to K.

Step 2
From (A4) applied to F and X,
 * $$\forall y \in F_t : y_p \subset \{ f = y_v \} \! $$
 * $$\forall z \in X_t : z_p \subset \{ x = z_v \} \! $$

combining them,
 * $$\forall y \in F_t, x \in X_t : y_p \subset \{ f = y_v \} \and x_p \subset \{ x = z_v \} \! $$

gives,
 * $$\forall y \in F_t, z \in X_t : y_p \cap x_p \subset \{ f = y_v \and x = z_v \} \! $$

as,
 * $$f = y_v \and x = z_v \implies f(x) = y_v(z_v) \!$$
 * $$\{f = y_v \and x = z_v\} \subset \{f(x) = y_v(z_v)\} \!$$

then
 * $$\forall y \in F_t, z \in X_t : y_p \cap x_p \subset \{ f(x) = y_v(z_v) \} \! $$

which is A4 applied to K.

Step 3
As $$K\!$$ satisfies the definition of a Value Set now prove that it equals $$F(X)\!$$. Let $$F = V(f)\!$$ and $$X = V(x)\!$$. Then $$F(X) = V(f)(V(x))\!$$.

From A6,
 * $$V(f(x)) = V(f)(V(x)) \!$$

From A1,
 * $$f \in S(V(f))\!$$
 * $$x \in S(V(x))\!$$

which implies,
 * $$f(x) \in \{y \in S(V(f))\and z \in S(V(x)):y(z)\} \!$$

so,
 * $$S(V(f(x)) = \{y \in S(V(f))\and z \in S(V(x)):y(z)\} \!$$

but also,
 * $$S(K) = S(v(\{f\in F_t, x\in X_t: f_v(x_v)::f_p\cap x_p\})) = \{y \in S(F)\and z \in S(X):y(z) \!$$

so from A2,
 * $$S(V(f(x)) = S(K) \implies V(f(x)) = K \!$$

so,
 * $$V(f(x))= V(f)(V(x)) = F(X) = K \!$$

Function of Vector of Parameters
If x is a vector, and X is a vector of Value Sets (representing multiple parameters) then,


 * (B4) $$F(X) = v(\{ f(x_v)::\bigcap_{j \in J} x_p[j] : x\in X \}) \!$$

Proof
Use induction. Firstly B1 gives the case for a scalar (n = 1).

Assume that B4 holds for a vector of size n then prove it holds for a vector of size n+1. Start with,


 * (B3) $$g(X, Y) = v(\{ g(x_v, y_v)::x_p\cap y_p : x\in X \and y\in Y \})\!$$

Let $$X\!$$ be a scalar and $$Y = Y(n)\!$$ is a vector of size n. Also let $$x\in X \and y(n) \in Y(n)\!$$.

Let,
 * $$y(n+1)_v = [x_v|y(n)_v] \!$$
 * $$y(n+1)_p = [x_p|y(n)_p] \!$$

then,
 * $$\bigcap_{j \in [1..n+1]} y(n+1)_p[j] = x_p \cap \bigcap_{j \in [1..n]} y(n)_p[j] \!$$

and,
 * $$y(n+1) \in Y(n+1) = x\in X \and y(n)\in Y(n) \!$$

Also let,
 * $$g(X, Y(n)) = h([X|Y(n)] = h(Y(n+1)) \!$$
 * $$g(x_v, y(n)_v) = h([x_v|y(n)_v]) = h(y(n+1)_v]) \!$$

in,


 * (B3) $$g(X, Y(n)) = v(\{ g(x_v, y(n)_v)::x_p\cap y(n)_p : x\in X \and y(n)\in Y(n) \})\!$$

becomes,


 * $$h(Y(n+1)) = v(\{ h(y(n+1)_v)::y(n+1)_p : y(n+1) \in Y(n+1) \})\!$$

Which is B4 for n+1, so by induction B4 holds for any number of parameters >= 1.

Assertions on Value Sets

 * (C1) $$x_V \implies \forall z \in V(x)_t: z_v \or (z_p = \empty) \!$$

Proof
From A5

$$V(x) \iff x \!$$

And, $$x \in \{ true \} \! $$.

Using,
 * (A1) $$x \in \{ z \in {V(x)}_t \and z_p \ne \empty : z_v \} \! $$

gives,
 * $$\{ z \in V(x)_t \and z_p \ne \empty : z_v \} = \{ true \} \! $$

and,
 * $$\forall z \in V(x)_t \and z_p \ne \empty : z_v = true \! $$

or,
 * $$\forall z \in V(x)_t: z_v \or z_p = \empty \! $$

Pruning Empty Worlds

 * (C2) $$X = v(\{x \in X \and x_p \neq \empty : x \}) \!$$


 * $$S(X) = \{ z \in X_t \and z_p \ne \empty : z_v \} \!$$

also,
 * $$S(v(\{x \in X \and x_p \neq \empty : x \})) = \{ z \in \{x \in X \and x_p \neq \empty : x \} \and z_p \ne \empty : z_v \} \!$$

which simplifies to,
 * $$S(v(\{x \in X \and x_p \neq \empty : x \})) = \{ z \in X_t \and z_p \ne \empty : z_v \} \!$$

so using A2,
 * $$S(X) = S(v(\{x \in X \and x_p \neq \empty : x \})) \implies X = v(\{x \in X \and x_p \neq \empty : x \}) \!$$

Resolution
The standard form for Possible Worlds is defined as,
 * $$X = f(Y, Z) \! $$
 * $$\forall x_p \in X, y_p \in Y: x_p \subset y_p or x_p \cap y_p = \empty \! $$

Then the resolution rule is,

If $$X \!$$, $$Y \!$$ and $$Z \!$$ are a Value Sets, where
 * $$X = f(Y, Z) \! $$

then,
 * $$(y_p \in S(Y) \and (\forall x_p \in S(X): x_p \not \subset y_p )) \implies y_p = \empty \! $$

Proof

 * $$x_p \subset y_p \or x_p \cap y_p = \empty $$

therefore,
 * $$\forall x_p \in S(X): x_p \not \subset y_p )) \implies \! $$
 * $$\forall x_p \in S(X): x_p \cap y_p = \empty )) \implies \! $$
 * $$\bigcup_{x_p \in S(X)} x_p \cap y_p = \empty \! $$

but,
 * $$\bigcup_{x_p \in S(X)} x_p \cap y_p = (\bigcup_{x_p \in S(X)} x_p) \cap y_p = \alpha \cap y_p = y_p\! $$

so,
 * $$y_p = \empty \! $$

Links

 * Symbolic Logic:Programming:Value Set
 * Symbolic Logic:Programming:Value Set Programming
 * Symbolic Logic:Programming
 * Intelligence and Reasoning