How to prove in K $ vdash_{p rightarrow lozenge square p} (square p rightarrow lozenge p)$?
$begingroup$
Let us denote $C$ the modal system obtained by adding the axiom $ alpha rightarrow lozenge square alpha$ to the axiom $K$ and all the propositional tautologies.
As said in the title, I'm looking for a K-proof for $$ vdash_C (square p rightarrow lozenge p),$$ but also for
$$vdash_C (lozenge p rightarrow square p).$$
I know these proofs should exist since a Kripke frame verifies the formula $p rightarrow lozenge square p$ if and only if it verifies $$ (1) forall x ( exists y : (x mathrel{R} y) text{ and } (y mathrel{R} z Rightarrow z = x) ).$$
It particular $(1)$ implies that the frame is serial ( i.e. $forall x (exists y : x mathrel{R} y)$), that is $ square p rightarrow lozenge p$ is true, and that the frame is functional ( i.e. ($x mathrel{R} y text{ and } x mathrel{R} z) Rightarrow y =z$), that is $lozenge p rightarrow square p$ is true.
Thank you for any answer !
logic modal-logic kripke-models
asked Dec 13 '18 at 15:28
L. De RudderL. De Rudder
335
$endgroup$
add a comment |
$begingroup$
Let us denote $C$ the modal system obtained by adding the axiom $ alpha rightarrow lozenge square alpha$ to the axiom $K$ and all the propositional tautologies.
As said in the title, I'm looking for a K-proof for $$ vdash_C (square p rightarrow lozenge p),$$ but also for
$$vdash_C (lozenge p rightarrow square p).$$
I know these proofs should exist since a Kripke frame verifies the formula $p rightarrow lozenge square p$ if and only if it verifies $$ (1) forall x ( exists y : (x mathrel{R} y) text{ and } (y mathrel{R} z Rightarrow z = x) ).$$
It particular $(1)$ implies that the frame is serial ( i.e. $forall x (exists y : x mathrel{R} y)$), that is $ square p rightarrow lozenge p$ is true, and that the frame is functional ( i.e. ($x mathrel{R} y text{ and } x mathrel{R} z) Rightarrow y =z$), that is $lozenge p rightarrow square p$ is true.
Thank you for any answer !
logic modal-logic kripke-models
asked Dec 13 '18 at 15:28
L. De RudderL. De Rudder
335
$endgroup$
add a comment |
$begingroup$
Let us denote $C$ the modal system obtained by adding the axiom $ alpha rightarrow lozenge square alpha$ to the axiom $K$ and all the propositional tautologies.
As said in the title, I'm looking for a K-proof for $$ vdash_C (square p rightarrow lozenge p),$$ but also for
$$vdash_C (lozenge p rightarrow square p).$$
I know these proofs should exist since a Kripke frame verifies the formula $p rightarrow lozenge square p$ if and only if it verifies $$ (1) forall x ( exists y : (x mathrel{R} y) text{ and } (y mathrel{R} z Rightarrow z = x) ).$$
It particular $(1)$ implies that the frame is serial ( i.e. $forall x (exists y : x mathrel{R} y)$), that is $ square p rightarrow lozenge p$ is true, and that the frame is functional ( i.e. ($x mathrel{R} y text{ and } x mathrel{R} z) Rightarrow y =z$), that is $lozenge p rightarrow square p$ is true.
Thank you for any answer !
logic modal-logic kripke-models
asked Dec 13 '18 at 15:28
L. De RudderL. De Rudder
335
$endgroup$
Let us denote $C$ the modal system obtained by adding the axiom $ alpha rightarrow lozenge square alpha$ to the axiom $K$ and all the propositional tautologies.
As said in the title, I'm looking for a K-proof for $$ vdash_C (square p rightarrow lozenge p),$$ but also for
$$vdash_C (lozenge p rightarrow square p).$$
I know these proofs should exist since a Kripke frame verifies the formula $p rightarrow lozenge square p$ if and only if it verifies $$ (1) forall x ( exists y : (x mathrel{R} y) text{ and } (y mathrel{R} z Rightarrow z = x) ).$$
It particular $(1)$ implies that the frame is serial ( i.e. $forall x (exists y : x mathrel{R} y)$), that is $ square p rightarrow lozenge p$ is true, and that the frame is functional ( i.e. ($x mathrel{R} y text{ and } x mathrel{R} z) Rightarrow y =z$), that is $lozenge p rightarrow square p$ is true.
Thank you for any answer !
logic modal-logic kripke-models
logic modal-logic kripke-models
asked Dec 13 '18 at 15:28
L. De RudderL. De Rudder
335
asked Dec 13 '18 at 15:28
L. De RudderL. De Rudder
335
edited Dec 14 '18 at 10:59
L. De Rudder
asked Dec 13 '18 at 15:28
L. De RudderL. De Rudder
335
asked Dec 13 '18 at 15:28
L. De RudderL. De Rudder
335
asked Dec 13 '18 at 15:28
L. De RudderL. De Rudder
335
335
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(For the first): Consider the set up where $neg p$ holds at the actual world and where no world is accessible from that world. Then the lhs of your sequent evaluates as true, and the right false ... (since $Box p$ is vacuously true, and $Diamond p$ is false).
(For the second): Consider the two-world set-up where in the actual self-accessible world $neg p$, and there is one other accessible world where $p$. The lhs of your sequent evaluates as true, and the right false ...
So something is amiss there? Neither is a valid K-sequent so neither is K-provable.
answered Dec 13 '18 at 15:51
Peter SmithPeter Smith
40.9k340120
$endgroup$
$begingroup$
I might be mistaken, but I think we can't deduct $p$ from $square p$ unless we are working with reflexive Kripke frames, which is not the case in general.
$endgroup$
– L. De Rudder
Dec 13 '18 at 16:15
$begingroup$
Oops, getting my K's and T's entangled!
$endgroup$
– Peter Smith
Dec 13 '18 at 16:58
2
$begingroup$
But I've given you, in each case, a K-model where the l.h. wff is true and the r.h. wff false. So (by soundness for the K proof system) the l.h. doesn't K-prove the r.h. wff. The implications aren't true.
$endgroup$
– Peter Smith
Dec 13 '18 at 20:32
2
$begingroup$
Agree on nomenclature of en.wikipedia.org/wiki/Kripke_semantics Then K is sound w.r.t. the inferences valid in every K frame, so in particular if $alpha vdash_K beta$, then given any K frame, and any valuation of the relevant atoms in that frame, if $alpha$ comes out true, so does $beta$. So it suffices to refute $alpha vdash_K beta$ to find one K frame and one valuation on that frame which makes $alpha$ true and $beta$ false. No?
$endgroup$
– Peter Smith
Dec 13 '18 at 23:14
1
$begingroup$
I'd have said "Let C be the modal system you get by adding to K all axioms of the form $alpha to DiamondBoxalpha$. Then how do we show $vdash_C Box p to Diamond p$, etc." [I've plucked the label "C" out of the air -- I don't know off-hand if there is a label already in use.]
$endgroup$
– Peter Smith
Dec 14 '18 at 10:42
|
show 8 more comments
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1 Answer
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$begingroup$
(For the first): Consider the set up where $neg p$ holds at the actual world and where no world is accessible from that world. Then the lhs of your sequent evaluates as true, and the right false ... (since $Box p$ is vacuously true, and $Diamond p$ is false).
(For the second): Consider the two-world set-up where in the actual self-accessible world $neg p$, and there is one other accessible world where $p$. The lhs of your sequent evaluates as true, and the right false ...
So something is amiss there? Neither is a valid K-sequent so neither is K-provable.
answered Dec 13 '18 at 15:51
Peter SmithPeter Smith
40.9k340120
$endgroup$
$begingroup$
I might be mistaken, but I think we can't deduct $p$ from $square p$ unless we are working with reflexive Kripke frames, which is not the case in general.
$endgroup$
– L. De Rudder
Dec 13 '18 at 16:15
$begingroup$
Oops, getting my K's and T's entangled!
$endgroup$
– Peter Smith
Dec 13 '18 at 16:58
2
$begingroup$
But I've given you, in each case, a K-model where the l.h. wff is true and the r.h. wff false. So (by soundness for the K proof system) the l.h. doesn't K-prove the r.h. wff. The implications aren't true.
$endgroup$
– Peter Smith
Dec 13 '18 at 20:32
2
$begingroup$
Agree on nomenclature of en.wikipedia.org/wiki/Kripke_semantics Then K is sound w.r.t. the inferences valid in every K frame, so in particular if $alpha vdash_K beta$, then given any K frame, and any valuation of the relevant atoms in that frame, if $alpha$ comes out true, so does $beta$. So it suffices to refute $alpha vdash_K beta$ to find one K frame and one valuation on that frame which makes $alpha$ true and $beta$ false. No?
$endgroup$
– Peter Smith
Dec 13 '18 at 23:14
1
$begingroup$
I'd have said "Let C be the modal system you get by adding to K all axioms of the form $alpha to DiamondBoxalpha$. Then how do we show $vdash_C Box p to Diamond p$, etc." [I've plucked the label "C" out of the air -- I don't know off-hand if there is a label already in use.]
$endgroup$
– Peter Smith
Dec 14 '18 at 10:42
|
show 8 more comments
$begingroup$
(For the first): Consider the set up where $neg p$ holds at the actual world and where no world is accessible from that world. Then the lhs of your sequent evaluates as true, and the right false ... (since $Box p$ is vacuously true, and $Diamond p$ is false).
(For the second): Consider the two-world set-up where in the actual self-accessible world $neg p$, and there is one other accessible world where $p$. The lhs of your sequent evaluates as true, and the right false ...
So something is amiss there? Neither is a valid K-sequent so neither is K-provable.
answered Dec 13 '18 at 15:51
Peter SmithPeter Smith
40.9k340120
$endgroup$
$begingroup$
I might be mistaken, but I think we can't deduct $p$ from $square p$ unless we are working with reflexive Kripke frames, which is not the case in general.
$endgroup$
– L. De Rudder
Dec 13 '18 at 16:15
$begingroup$
Oops, getting my K's and T's entangled!
$endgroup$
– Peter Smith
Dec 13 '18 at 16:58
2
$begingroup$
But I've given you, in each case, a K-model where the l.h. wff is true and the r.h. wff false. So (by soundness for the K proof system) the l.h. doesn't K-prove the r.h. wff. The implications aren't true.
$endgroup$
– Peter Smith
Dec 13 '18 at 20:32
2
$begingroup$
Agree on nomenclature of en.wikipedia.org/wiki/Kripke_semantics Then K is sound w.r.t. the inferences valid in every K frame, so in particular if $alpha vdash_K beta$, then given any K frame, and any valuation of the relevant atoms in that frame, if $alpha$ comes out true, so does $beta$. So it suffices to refute $alpha vdash_K beta$ to find one K frame and one valuation on that frame which makes $alpha$ true and $beta$ false. No?
$endgroup$
– Peter Smith
Dec 13 '18 at 23:14
1
$begingroup$
I'd have said "Let C be the modal system you get by adding to K all axioms of the form $alpha to DiamondBoxalpha$. Then how do we show $vdash_C Box p to Diamond p$, etc." [I've plucked the label "C" out of the air -- I don't know off-hand if there is a label already in use.]
$endgroup$
– Peter Smith
Dec 14 '18 at 10:42
|
show 8 more comments
$begingroup$
(For the first): Consider the set up where $neg p$ holds at the actual world and where no world is accessible from that world. Then the lhs of your sequent evaluates as true, and the right false ... (since $Box p$ is vacuously true, and $Diamond p$ is false).
(For the second): Consider the two-world set-up where in the actual self-accessible world $neg p$, and there is one other accessible world where $p$. The lhs of your sequent evaluates as true, and the right false ...
So something is amiss there? Neither is a valid K-sequent so neither is K-provable.
answered Dec 13 '18 at 15:51
Peter SmithPeter Smith
40.9k340120
$endgroup$
(For the first): Consider the set up where $neg p$ holds at the actual world and where no world is accessible from that world. Then the lhs of your sequent evaluates as true, and the right false ... (since $Box p$ is vacuously true, and $Diamond p$ is false).
(For the second): Consider the two-world set-up where in the actual self-accessible world $neg p$, and there is one other accessible world where $p$. The lhs of your sequent evaluates as true, and the right false ...
So something is amiss there? Neither is a valid K-sequent so neither is K-provable.
answered Dec 13 '18 at 15:51
Peter SmithPeter Smith
40.9k340120
edited Dec 14 '18 at 8:50
answered Dec 13 '18 at 15:51
Peter SmithPeter Smith
40.9k340120
answered Dec 13 '18 at 15:51
Peter SmithPeter Smith
40.9k340120
answered Dec 13 '18 at 15:51
Peter SmithPeter Smith
40.9k340120
40.9k340120
$begingroup$
I might be mistaken, but I think we can't deduct $p$ from $square p$ unless we are working with reflexive Kripke frames, which is not the case in general.
$endgroup$
– L. De Rudder
Dec 13 '18 at 16:15
$begingroup$
Oops, getting my K's and T's entangled!
$endgroup$
– Peter Smith
Dec 13 '18 at 16:58
2
$begingroup$
But I've given you, in each case, a K-model where the l.h. wff is true and the r.h. wff false. So (by soundness for the K proof system) the l.h. doesn't K-prove the r.h. wff. The implications aren't true.
$endgroup$
– Peter Smith
Dec 13 '18 at 20:32
2
$begingroup$
Agree on nomenclature of en.wikipedia.org/wiki/Kripke_semantics Then K is sound w.r.t. the inferences valid in every K frame, so in particular if $alpha vdash_K beta$, then given any K frame, and any valuation of the relevant atoms in that frame, if $alpha$ comes out true, so does $beta$. So it suffices to refute $alpha vdash_K beta$ to find one K frame and one valuation on that frame which makes $alpha$ true and $beta$ false. No?
$endgroup$
– Peter Smith
Dec 13 '18 at 23:14
1
$begingroup$
I'd have said "Let C be the modal system you get by adding to K all axioms of the form $alpha to DiamondBoxalpha$. Then how do we show $vdash_C Box p to Diamond p$, etc." [I've plucked the label "C" out of the air -- I don't know off-hand if there is a label already in use.]
$endgroup$
– Peter Smith
Dec 14 '18 at 10:42
|
show 8 more comments
$begingroup$
I might be mistaken, but I think we can't deduct $p$ from $square p$ unless we are working with reflexive Kripke frames, which is not the case in general.
$endgroup$
– L. De Rudder
Dec 13 '18 at 16:15
$begingroup$
Oops, getting my K's and T's entangled!
$endgroup$
– Peter Smith
Dec 13 '18 at 16:58
2
$begingroup$
But I've given you, in each case, a K-model where the l.h. wff is true and the r.h. wff false. So (by soundness for the K proof system) the l.h. doesn't K-prove the r.h. wff. The implications aren't true.
$endgroup$
– Peter Smith
Dec 13 '18 at 20:32
2
$begingroup$
Agree on nomenclature of en.wikipedia.org/wiki/Kripke_semantics Then K is sound w.r.t. the inferences valid in every K frame, so in particular if $alpha vdash_K beta$, then given any K frame, and any valuation of the relevant atoms in that frame, if $alpha$ comes out true, so does $beta$. So it suffices to refute $alpha vdash_K beta$ to find one K frame and one valuation on that frame which makes $alpha$ true and $beta$ false. No?
$endgroup$
– Peter Smith
Dec 13 '18 at 23:14
1
$begingroup$
I'd have said "Let C be the modal system you get by adding to K all axioms of the form $alpha to DiamondBoxalpha$. Then how do we show $vdash_C Box p to Diamond p$, etc." [I've plucked the label "C" out of the air -- I don't know off-hand if there is a label already in use.]
$endgroup$
– Peter Smith
Dec 14 '18 at 10:42
$begingroup$
I might be mistaken, but I think we can't deduct $p$ from $square p$ unless we are working with reflexive Kripke frames, which is not the case in general.
$endgroup$
– L. De Rudder
Dec 13 '18 at 16:15
$begingroup$
I might be mistaken, but I think we can't deduct $p$ from $square p$ unless we are working with reflexive Kripke frames, which is not the case in general.
$endgroup$
– L. De Rudder
Dec 13 '18 at 16:15
$begingroup$
Oops, getting my K's and T's entangled!
$endgroup$
– Peter Smith
Dec 13 '18 at 16:58
$begingroup$
Oops, getting my K's and T's entangled!
$endgroup$
– Peter Smith
Dec 13 '18 at 16:58
2
2
$begingroup$
But I've given you, in each case, a K-model where the l.h. wff is true and the r.h. wff false. So (by soundness for the K proof system) the l.h. doesn't K-prove the r.h. wff. The implications aren't true.
$endgroup$
– Peter Smith
Dec 13 '18 at 20:32
$begingroup$
But I've given you, in each case, a K-model where the l.h. wff is true and the r.h. wff false. So (by soundness for the K proof system) the l.h. doesn't K-prove the r.h. wff. The implications aren't true.
$endgroup$
– Peter Smith
Dec 13 '18 at 20:32
2
2
$begingroup$
Agree on nomenclature of en.wikipedia.org/wiki/Kripke_semantics Then K is sound w.r.t. the inferences valid in every K frame, so in particular if $alpha vdash_K beta$, then given any K frame, and any valuation of the relevant atoms in that frame, if $alpha$ comes out true, so does $beta$. So it suffices to refute $alpha vdash_K beta$ to find one K frame and one valuation on that frame which makes $alpha$ true and $beta$ false. No?
$endgroup$
– Peter Smith
Dec 13 '18 at 23:14
$begingroup$
Agree on nomenclature of en.wikipedia.org/wiki/Kripke_semantics Then K is sound w.r.t. the inferences valid in every K frame, so in particular if $alpha vdash_K beta$, then given any K frame, and any valuation of the relevant atoms in that frame, if $alpha$ comes out true, so does $beta$. So it suffices to refute $alpha vdash_K beta$ to find one K frame and one valuation on that frame which makes $alpha$ true and $beta$ false. No?
$endgroup$
– Peter Smith
Dec 13 '18 at 23:14
1
1
$begingroup$
I'd have said "Let C be the modal system you get by adding to K all axioms of the form $alpha to DiamondBoxalpha$. Then how do we show $vdash_C Box p to Diamond p$, etc." [I've plucked the label "C" out of the air -- I don't know off-hand if there is a label already in use.]
$endgroup$
– Peter Smith
Dec 14 '18 at 10:42
$begingroup$
I'd have said "Let C be the modal system you get by adding to K all axioms of the form $alpha to DiamondBoxalpha$. Then how do we show $vdash_C Box p to Diamond p$, etc." [I've plucked the label "C" out of the air -- I don't know off-hand if there is a label already in use.]
$endgroup$
– Peter Smith
Dec 14 '18 at 10:42
|
show 8 more comments
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