Soundness of Single Step Tableau Rules

Example 6.2.1.

Below we show a closed systematic ^LCK-tableau for X = ^f2(p ^! q);^:(²p ^{! 2}q)^g. We assume that A ^! B is written as

:(A^{^:}B) and that ^:(A ^! B) is rewritten and simplied to A^{^:}B. We use \a", \s" and \f" for

a

^{wake, a}

s

^{leep and}

f

inished respectively. The notation s/a indicates that the formula was asleep but was woken up during the stage.

Systematic ^LCK-tableau forX =^f2(p^!q);^:(²p^!2q)^g

Extant Tableau at Marks at

Stage End Stage End

1 2 3 4 5 6 7 8 9 1 :: ²(p^!q) a s s s s/a s s s s 1 :: ^:(²p^!2q) a a f f f f f f f

1 :: ²p a s s/a a s s s

1 :: ^:2q a a f f f f f

1:1 :: ^:q a a a f f

1:1 :: p^!q a a a f

1:1 :: p a a a

1:1 ::^:p a

1:1 ::^::q a

(closed) (closed)

Example 6.2.2.

The formula (²³p)^{^}(³p) can be written in primitive notation as (^2:2:p)^{^}(^:2:p). As the reader can verify, the systematic

S4

-tableau forX =^f(^2:2:p)^{^}(^:2:p)^gneither terminates nor closes.

Denition of where and are nonempty and drawn from a strongly generated set of labels ? with root= 1 Logics for all; ²?, is

L

-accessible from i

K

=:nfor somen1

KT

^{=:nor =}

KB

=:nor =:m

K4

^{=: andjj1}

K5

=:nor (^jj2 and^jj2)

K45

^{( =:andjj1) or (jj2 andjj2)}

KD K

-condition or ( is a

K

-deadend and=)

KDB KB

-condition or (^j?^j= 1 and= = 1)

KD4 K4

-condition or ( is a

K

-deadend and =)

KD5 K5

-condition or (^j?^j= 1 and = = 1)

KD45 K45

-condition or (^j?^j= 1 and= = 1)

KB4

^j?^j2

B

^{= or =:n or=:m}

S4

( =:and^jj1) or ( =)

S5

^j?j1

Fig. 18.

Denition of

L

-accessibility.

assumption that the root= 1 simplies the conditions for

L

-accessibility, but there is still a slight complication for the serial logics.

For any nonserial logic

L

¹we say thatis an

L

¹

-deadend

if there is no that is

L

¹-accessible from. Now we can express the seriality condition for the serial counterpart

L

=

L

¹

D

by demanding that all

L

¹-deadends be reexive. In particular, we say that ²? is a

K

-deadend if no label in

? is a simple extension of . In Figure 18 we have computed the forms of the

L

¹-deadends and added an extra condition to make them reexive for each logic

L

¹

D

. The notation^j?^j means the number of labels in ?.

We leave it to the reader to generalise these conditions to account for the case where is an arbitrary label. Note that the conditions on

L

-accessibility in Figure 18 and the conditions on -accessibility in the nite-

L

-frames of Figure 13 on page 67 are closely related. We return to this point later.

But rst we relate

L

-accessibility to the

L

-frames of Figure 4 on page 9.

Theorem 6.3.1.

If ? is a strongly generated set of labels with root= 1 then^F =^h?;ⁱ is an

L

-frame.

Proof:

It is obvious that

KT

-accessibility,

K4

-accessibility and

KB

-accessibility forces^Fto be respectively reexive, transitive and symmetric.

We consider only the case for

K45

in detail.

We have to show that

K45

-accessibility forces ^F to be euclidean and

transitive.

K45

-accessibilityis euclidean if ⁰¹ and⁰²implies ¹², where

K45

-accessibilityis dened as:

i ( =: and^jj1) or (^jj2 and^jj2) By substitution we get:

Hypotheses Expanded Hypothesis

⁰¹ (¹=⁰:¹ and^j1j1) or (^j0j2 and ^j1j2)

and and

⁰² (²=⁰:² and^j2j1) or (^j0j2 and ^j2j2)

Goal Expanded Goal

¹² (²=¹:³ and^j3j1) or (^j1j2 and^j2j2) Now, we know that ⁰ is nonempty, hence ^j0j1. But this together with (¹ =⁰:¹ and^j1j 1) in the left disjunct of the rst hypothesis immediately gives ^j1j 2. Thus both disjuncts of the rst line of the hypothesis imply^j1j2.

Similarly, ^j0j 1 together with (² =⁰:² and^j2j1) in the left disjunct of the second hypothesis gives ^j2j 2. Thus both disjuncts of the second hypothesis imply ^j2j2).

And the conjunction of these two gives the second disjunct of the goal showing that

K45

-accessibility relationis indeed euclidean.

To show that

K45

-accessibility is also transitive, we must show that ⁰¹ and ¹² implies⁰². The same expansions can be used but the roles of hypotheses and goal are slightly altered. The argument is almost identical, except for one subcase which relies on the fact that

j^0j= 1 implies⁰= 1.

Let ^X be a strongly generated set of labelled formulae, let lab(^X) be the set of labels that appear in^X and let^M=^hW;R;Vⁱbe some

L

-model where

L

is any one of the 15 distinct basic normal modal logics obtainable by adding any combination of the axioms T, D, B, 4 and 5 to logic

K

^.

Call a world in^M

idealisable

i it has anR-successor in^M.

An

L-interpretation of

(a strongly generated set of labelled formulae)

X

in

^M is a mapping I : lab(^X)^7! W that satises: if and I() is idealisable then I()RI(), where is the appropriate

L

-accessibility relation from Figure 18 [Fit83].

A strongly generated set^X of labelled formulae is

L-satisable under

the L-interpretation

^{I if I() j= A for each :: A in X; and is}

L-satisable

if it is

L

-satisable under some

L

-interpretation. A branch of a labelled tableau is

L

-satisable if the set of labelled formulae on it is

L

-satisable, and a tableau is

L

-satisable if some branch of the tableau is

L

-satisable.

Proposition 6.3.2.

The set of labelled formulaelab(^B) from any branch

B of a labelled tableau is a strongly generated set.

Proof:

By the fact that the initial label is always= 1, and the fact that the only new labels that may be created are labels of the form :n, n1, which are all simple extensions of some²lab(^B).

We now prove soundness of some of the rules leaving the others to the reader. Since the systematic procedure updates all branches that pass through the chosen formula, the soundness theorem states the following: if a tableau^T is

L

-satisable and we apply rule (l) to get tableau^T0, then

T

0 is also

L

-satisable. Since every rule has at most two denominators, a rule can cause a given branch to split into at most branches. Consequently we have to prove that if a branch^Bis

L

-satisable, and applying rule (l) causes it to be updated into branches ^C and ^D, then at least one of the new branches is also

L

-satisable.

Soundness of

^(l)

for L-frames:

^{SupposeBis an}

L

-satisable branch and that we apply the (l) rule to some awake :: ^:2P on^B to obtain branch^C containingn::^:P where nis a simple extension of that is new to^B. We have to show that^C is

L

-satisable.

Since^Bis

L

-satisable, there is some

L

^{-modelM=hW;R;Viand some}

L

-interpretationIin^Msuch thatI()²W andI()^j=^:2P. HenceI() is idealisable as there is somew²W withI()Rw andw^j=^:P. Sincen is new, it does not appear in^Band hence has no image underI. ExtendI by puttingI(n) =w. We then haven,I()RI(n), andI(n)^j=^:P meaning that^C is indeed

L

-satisable under the extendedI in^M.

Soundness of

(l4^d)

for K5-frames:

Suppose^B is a

K5

-satisable branch and that we apply the (l4^d) rule to somen::²P to get a branch

C containingnm::²P. We have to show that^Cis also

K5

-satisable.

Since^Bis

K5

-satisable and the labelsnandnmmust already exist on^B, there is some

K5

-model^M=^hW;R;Vⁱand some

K5

-interpretation Iin^Msuch thatI(n)²W,I(nm)²W andI(n)^j=²P. The labeln can exist on^Bonly ifalso exists on^Bsince^Bis strongly generated. Hence there is someI()²W. The congurationnnmimmediately im-pliesI()RI(n)RI(nm) by the denition ofI. BecauseRis euclidean we know that I(n)RI(n); that is I(n) is reexive. ThenI(n)RI(nm) and I(n)RI(n) gives I(nm)RI(n). Hence I(nm) ^j= ³²P. Eu-clidean frames must validate axiom 5 (³²A^!2A) henceI(nm)^j=²P. We have not alteredI in any way, so by denition,^Cis

K5

-satisable under I in^M.

Soundness of

(l5)

for K5-frames:

Suppose ^B is a

K5

-satisable

branch and that we apply the (l5) rule to some 1:n::²P to get a branch

C containing 1 ::²²P. We have to show that^C is also

K5

-satisable.

As before there is some

K5

^{-model M = hW;R;Vi and some}

K5

-interpretation I in ^M such that I(1:n) ² W and I(1:n) ^j= ²P. Since 1 is used on^Band 11:n, there must be someI(1)²W withI(1)RI(1:n).

Now suppose for a contradiction thatI(1)^j=^:22P; then there is some w²W such thatI(1)Rw andw^j=^:2P, which in turn implies that there is some w^{0 2} W such that wRw⁰ and w^{0 j}= ^:P. Since R is euclidean, I(1)RI(1:n) andI(1)RwgiveswRI(1:n), and thenwRw⁰ givesI(1:n)Rw⁰. But then I(1:n)^j=²P impliesw^{0 j}=P; contradiction. HenceI(1)^j=²²P and^C is

K5

-satisable underI in ^M.

Theorem 6.3.3.

If the systematic tableau for X closes then X is

L

-unsatisable.

Proof:

For a contradiction, suppose the tableau for X is closed and thatX is

L

-satisable. The latter means that there is some

L

-model^M=

hW;R;Vⁱ and some worldw²W such thatw ^j=X. Our tableau begins with nodes 1 ::Ai, for eachAi ²X so dene an

L

-interpretationI in^M such thatI(1) =w. Then the initial tableau comprising the linear sequence of these nodes 1 :: Ai is

L

-satisable (under I in ^M). Since each of our tableau rules is sound, any tableau obtained from this initial tableau by these rules is also

L

-satisable. Hence our tableau is

L

-satisable.

Suppose^Bis some branch of this closed tableau. Then^Bitself is closed and hence contains some labelled formula ::P and also contains::^:P. Now any

L

-interpretation I⁰ for ^B in any

L

^{-model M0} would entail that I⁰()^j=P and also that I⁰()^j=^:P, which is clearly impossible. Hence

B is not

L

-satisable. Since ^Bwas an arbitrary branch this must be true for all branches of this closed tableau. Then, by denition, our tableau is not

L

-satisable. Contradiction, hence if the tableau for X closes thenX is

L

-unsatisable.

Corollary 6.3.4 (soundness).

If the systematic tableau for^f:A^gis closed thenA is

L

-valid.

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