Why is this formula valid in PDL
$begingroup$
In the book modal logic for open minds by johan van benthem there is on page 161 a statement that the sentence $langle (Rlor S)^* rangle phi$ is equivalent to the sentence $langle (R^* ; S^*)^* rangle phi$
(* means iteration and ; means composition here)
So:
$langle (Rlor S)^* rangle phi equiv langle (R^* ; S^*)^* rangle phi$
Why is this equivalent to each other?
logic modal-logic
asked Jan 18 at 17:08
jennifer ruursjennifer ruurs
445
$endgroup$
add a comment |
$begingroup$
In the book modal logic for open minds by johan van benthem there is on page 161 a statement that the sentence $langle (Rlor S)^* rangle phi$ is equivalent to the sentence $langle (R^* ; S^*)^* rangle phi$
(* means iteration and ; means composition here)
So:
$langle (Rlor S)^* rangle phi equiv langle (R^* ; S^*)^* rangle phi$
Why is this equivalent to each other?
logic modal-logic
asked Jan 18 at 17:08
jennifer ruursjennifer ruurs
445
$endgroup$
add a comment |
$begingroup$
In the book modal logic for open minds by johan van benthem there is on page 161 a statement that the sentence $langle (Rlor S)^* rangle phi$ is equivalent to the sentence $langle (R^* ; S^*)^* rangle phi$
(* means iteration and ; means composition here)
So:
$langle (Rlor S)^* rangle phi equiv langle (R^* ; S^*)^* rangle phi$
Why is this equivalent to each other?
logic modal-logic
asked Jan 18 at 17:08
jennifer ruursjennifer ruurs
445
$endgroup$
In the book modal logic for open minds by johan van benthem there is on page 161 a statement that the sentence $langle (Rlor S)^* rangle phi$ is equivalent to the sentence $langle (R^* ; S^*)^* rangle phi$
(* means iteration and ; means composition here)
So:
$langle (Rlor S)^* rangle phi equiv langle (R^* ; S^*)^* rangle phi$
Why is this equivalent to each other?
logic modal-logic
logic modal-logic
asked Jan 18 at 17:08
jennifer ruursjennifer ruurs
445
asked Jan 18 at 17:08
jennifer ruursjennifer ruurs
445
asked Jan 18 at 17:08
jennifer ruursjennifer ruurs
445
asked Jan 18 at 17:08
jennifer ruursjennifer ruurs
445
asked Jan 18 at 17:08
jennifer ruursjennifer ruurs
445
445
add a comment |
add a comment |
1 Answer
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$begingroup$
Suppose that for some arbitrary $M_w$ it holds that $M_w models langle (r cup s)^ast rangle varphi$. By the semantics this means that $exists v in W^M$ such that $M_{w,v} models (r cup s)^ast$ and $M_v models varphi$. Now let us consider $M_{w,v} models (r cup s)^ast$. According to the semantics, there is an $r cup s$ path (which may be empty) between $w$ and $v$. Expression $r cup s$ means that every step of the path may be either an $r$-step or an $s$-step. Examples of such paths are the empty path, $r$, $s,s$, $r,s,r,s,s,r$ and so on. Now we notice that the same paths can be described by the regular expression $(r^ast;s^ast)^ast$. Moreover, we could construct automata for recognising $(r cup s)^ast$ and $(r^ast;s^ast)^ast$, and see that they are equivalent. Therefore, we have that $M_{w,v} models (r^ast;s^ast)^ast$ and $M_v models varphi$, and hence $M_w models langle (r^ast;s^ast)^ast rangle varphi$.
answered Feb 15 at 20:03
Charles BronsonCharles Bronson
572213
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$begingroup$
Suppose that for some arbitrary $M_w$ it holds that $M_w models langle (r cup s)^ast rangle varphi$. By the semantics this means that $exists v in W^M$ such that $M_{w,v} models (r cup s)^ast$ and $M_v models varphi$. Now let us consider $M_{w,v} models (r cup s)^ast$. According to the semantics, there is an $r cup s$ path (which may be empty) between $w$ and $v$. Expression $r cup s$ means that every step of the path may be either an $r$-step or an $s$-step. Examples of such paths are the empty path, $r$, $s,s$, $r,s,r,s,s,r$ and so on. Now we notice that the same paths can be described by the regular expression $(r^ast;s^ast)^ast$. Moreover, we could construct automata for recognising $(r cup s)^ast$ and $(r^ast;s^ast)^ast$, and see that they are equivalent. Therefore, we have that $M_{w,v} models (r^ast;s^ast)^ast$ and $M_v models varphi$, and hence $M_w models langle (r^ast;s^ast)^ast rangle varphi$.
answered Feb 15 at 20:03
Charles BronsonCharles Bronson
572213
$endgroup$
add a comment |
$begingroup$
Suppose that for some arbitrary $M_w$ it holds that $M_w models langle (r cup s)^ast rangle varphi$. By the semantics this means that $exists v in W^M$ such that $M_{w,v} models (r cup s)^ast$ and $M_v models varphi$. Now let us consider $M_{w,v} models (r cup s)^ast$. According to the semantics, there is an $r cup s$ path (which may be empty) between $w$ and $v$. Expression $r cup s$ means that every step of the path may be either an $r$-step or an $s$-step. Examples of such paths are the empty path, $r$, $s,s$, $r,s,r,s,s,r$ and so on. Now we notice that the same paths can be described by the regular expression $(r^ast;s^ast)^ast$. Moreover, we could construct automata for recognising $(r cup s)^ast$ and $(r^ast;s^ast)^ast$, and see that they are equivalent. Therefore, we have that $M_{w,v} models (r^ast;s^ast)^ast$ and $M_v models varphi$, and hence $M_w models langle (r^ast;s^ast)^ast rangle varphi$.
answered Feb 15 at 20:03
Charles BronsonCharles Bronson
572213
$endgroup$
add a comment |
$begingroup$
Suppose that for some arbitrary $M_w$ it holds that $M_w models langle (r cup s)^ast rangle varphi$. By the semantics this means that $exists v in W^M$ such that $M_{w,v} models (r cup s)^ast$ and $M_v models varphi$. Now let us consider $M_{w,v} models (r cup s)^ast$. According to the semantics, there is an $r cup s$ path (which may be empty) between $w$ and $v$. Expression $r cup s$ means that every step of the path may be either an $r$-step or an $s$-step. Examples of such paths are the empty path, $r$, $s,s$, $r,s,r,s,s,r$ and so on. Now we notice that the same paths can be described by the regular expression $(r^ast;s^ast)^ast$. Moreover, we could construct automata for recognising $(r cup s)^ast$ and $(r^ast;s^ast)^ast$, and see that they are equivalent. Therefore, we have that $M_{w,v} models (r^ast;s^ast)^ast$ and $M_v models varphi$, and hence $M_w models langle (r^ast;s^ast)^ast rangle varphi$.
answered Feb 15 at 20:03
Charles BronsonCharles Bronson
572213
$endgroup$
Suppose that for some arbitrary $M_w$ it holds that $M_w models langle (r cup s)^ast rangle varphi$. By the semantics this means that $exists v in W^M$ such that $M_{w,v} models (r cup s)^ast$ and $M_v models varphi$. Now let us consider $M_{w,v} models (r cup s)^ast$. According to the semantics, there is an $r cup s$ path (which may be empty) between $w$ and $v$. Expression $r cup s$ means that every step of the path may be either an $r$-step or an $s$-step. Examples of such paths are the empty path, $r$, $s,s$, $r,s,r,s,s,r$ and so on. Now we notice that the same paths can be described by the regular expression $(r^ast;s^ast)^ast$. Moreover, we could construct automata for recognising $(r cup s)^ast$ and $(r^ast;s^ast)^ast$, and see that they are equivalent. Therefore, we have that $M_{w,v} models (r^ast;s^ast)^ast$ and $M_v models varphi$, and hence $M_w models langle (r^ast;s^ast)^ast rangle varphi$.
answered Feb 15 at 20:03
Charles BronsonCharles Bronson
572213
answered Feb 15 at 20:03
Charles BronsonCharles Bronson
572213
answered Feb 15 at 20:03
Charles BronsonCharles Bronson
572213
answered Feb 15 at 20:03
Charles BronsonCharles Bronson
572213
572213
add a comment |
add a comment |
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