Treatment of Ross' Paradox along Griceian lines

From

http://akira.ruc.dk/~mamobe/finalthesis

"The two following proof shows that Ross´ disjunctive paradox can be derived in both of the forms discussed by Ross himself."

From “Post the letter!”

we can infer both

“Burn the letter or post the letter”

and “you have to burn the letter or you have to send the letter”.

Q: You send the letter
P: You burn the letter

Proof 32.1) Proof 32.2)

1 OQ 1 OQ

1 ¬O(P∨Q) 1 ¬(OP∨OQ)

1.1 ¬(P∨Q) 1 ¬OQ

1.1 ¬P X

1.1 ¬Q

1.1 Q

X

"A defence is that a disjunctive obligation does not imply a conjunctive or choice-offering permission as it sometimes seems to do in ordinary language."

"If I am told that I either have burn the letter or have to send it, it could be taken to mean that I choose which one to do."

"But in SDL we cannot derive

PP∧PQ

from

OP∨OQ.

OP∨OQ is sometimes called an alternative-presenting obligation.
Tree 33)

1 OP∨OQ

1¬(PP∧PQ)

 

1¬ PP 1 ¬ PQ

1 ¬OP 1 ¬OQ

   

1 OP 1 OQ 1 OP 1 OQ

X 1.1 ¬P 1.2 ¬Q X

1.1 Q 1.2 P

"The open branches can be turned into counter models."

"We concentrate on the first open branch from the left."

"Let g be 1, 1.1. with 1ℜ1.1, 1.1ℜ1.1125, and with

1. 1╟Q

and

1.

1ǪP. It follows that
125 1.1 being related to itself gives it a point related to it and makes the frame serial.

"The full formal definition of the
frame 〈g, ℜ〉 is g={1, 1.1} ℜ={ 〈1,1.1 〉, 〈1.1,1.1 〉 }. So I take the prefixes themselves to be the “worlds” of the frame.

1╟ OQ and thus 1╟ OP∨OQ. It also follows that 1╟¬PP thus not(1╟ PP∧PQ) and so 1╟
¬(PP∧PQ). This is also a counter model to P(P∧Q) from OP∨OQ , since we have 1.1╟ ¬(P∧Q)
and so 1╟¬ P(P∧Q). Nor can P(P∧Q) or (PP∧PQ) be derived from O(P∨Q) in SDL. Nor do any of

the converses hold. I leave most of the remaining display of counter models to the interested reader.

Here I will just show an interesting special case. A free-choice permission is of the form (PP∧P¬P).

If P is that you smoke it means that you are permitted to smoke and permitted not to smoke. It is not
possible to derive that either P or ¬P are obligatory( that you have to smoke or not smoke) from
this.
Tree 34)

1 PP∧P¬P

1 ¬(OP∨O¬P)

1 ¬OP

1 ¬O¬P

1 PP

1 P¬P

1.1 P

1.2 ¬P

Tree 34) shows how OP and its contrary can both be false and how the subcontraries PP and P¬P
can both be true without any contradiction. This corresponds with the deontic square of opposition
above. A formal semantic counter model has 1.1╟ P, 1.2╟¬P and 1ℜ1.1, 1ℜ1.2, (plus 1.1ℜ1.1 and
1.2 ℜ1.2 to make it serial). Both the permissions are true and both the obligations are false in this
model, as the reader may verify.(¬P is true at 1.2. so P¬P is true at 1. Similarly for PP so
(PP∧P¬P) is true at 1, etc.).
The same considerations go for the permissive variant of Ross´ paradox from PP to P(P∨Q). It
does not follow that P and Q are both permitted. The right formalisation of a free choice permission
is PP∧PQ126.

"This concludes my discussion of Ross´ disjunctive paradox in the context of SDL."