# [FOM] negation vs ituitionism

Åke Persson ok.person at swipnet.se
Sat Mar 22 07:51:18 EST 2003

```Negation vs intuitionism
-------------------------

To describe and understand the problems with reduction ad absurbum and
double negation from intuitionistic point of view, it is necessary to go
outside bivalued contexts and distinguish and sharpen up concepts that are
collapsed together in bivalued contexts. In so called 'intuitionistic
logics' the rule of double negation elimination is rejected as the
troblusome rule. Using quantity relation logic we more exactly can adress
the problems with reduction ad absurbum to some properties of modal
contexts, that leaves the rule of double negation elimination untouched.
However, there is a long way to go, to first sharpen up some traditional
concepts before it is possible to right address and show the problem.

First we need to distinguish between negation of meaning and truth.

When we denies the mening of p we get ~p that is an other meaning, the
contrary meaning 'not-p'. Then we denies the meaning 'not-p' we get the
contrary meaning to 'not-p', which is p. When we denies the truth of p the
meaning p we denies does not change. Let's use the prefix T and F for true
and false and assume as in classical logic that truth and falseness is
contraries such as T and ~F means the same and F and ~T means the same. When
we assumes that p is false, we assumes Fp. This is a new mening that
includes the meaning of p but now with the additional meaning 'to be false'.
This new statement Fp has quite different properties than the statement of
~p. Then we denies Fp we get the contrary to the meaning Fp which is ~Fp,
i.e. 'p is not false', i.e. 'p is true', i.e. Tp which meaning differ from
p. Then we denies ~p we get the contrary to the meaning ~p which is p.

Next we need to distinguish between "not-1" and 0.

This point can not ever be discovered or understood inside a bivalued
context as the only two values T and F are represented by 1 and 0 and
'not-1' only can be 0. To understand the difference we need to look at a
range of numbers from 0 to 1 including them, i.e. the interval [0,1] or,
which is the same, [0%,100%]. Then p is true we also can understand this as
'p has the truthvalue 1', i.e. 'p = 1'. Then we denies this mening it is the
equality we denies, and we get 'p < 1 or p > 1' which means the same as 'p <
1' as 1 is the largest number in this context. When we again denies 'p < 1'
we get 'p = 1 or p > 1' which means the same as 'p = 1' . So, as 'p = 1'
means the same as Tp above '~(p = 1)', i.e. 'p < 1' must mean the same as
Fp. In [0,1] the difference between 'p < 1' and 'p = 0' is of great
importance. In bivalence these different writings colapses into each others.
For easier handling, let's use following definitions and semantics:

(1)    Tp    =df    (p = 1)        ; 'p is 100% true'
(2)    Fp    =df    (p < 1)        ; 'p is lower than 100% true'
(3)    Ep    =df    (p > 0)        ; 'p is higher than 0% true'
(4)    Up    =df    (p = 0)        ; 'p is 0% true'

As exampel of an applicable context: Think of an urn with 100 marbles and
the statement p:  'The marbles in the urn is black'. If all of them are
black, the statement is 100% true, if none are black it is 0% true. If there
are only 99 black mables the statement is false (but still nearly true).
This context has 101 possible values (the number of marbles + 1). If the urn
only had one marble, the context should be bivalued.

Next we need to right address a third alternative 'unknown'

In the above mentioned context the truthvalues of the statements p and ~p
together must be 'the hole', i.e. 100%. However, some marbles can remain
unknown of some reason, e.g. until all of them are out and visible. Now a
third alternative appears to make the sum of p and ~p lower than 100% and
the statement 'It is unknown if the marbles in the urn is black or not
black' is true to the difference.  Let us simplify and use the prefix '?' to
denote 'it is unknown...' and denote the truthvalues as relative occurences
as above with the prefix R. Now the sum of the three alternatives is:

(5)    Rp + R~p + R?p == 1

'Unknown' is now neither a new truth value between true and false nor a new
meaning between p and ~p, but is a new alternative that only concur of the
whole avilable truthvalue, 100%, not able to touch the contrarity of p and
~p or true and false. To introduce 'unknown' as a *truth value* beside or
between 'true' and 'false' introduces only big philosophical and unsolvable
troubles.

Next we need to apply this three-alternativs-context to the four definitions
above

As no marble can satisfy more than one of the statements p, ~p or ?p, the
sum (5) also can be expressed as (the proofs are excluded here):

(5.1)    R(p v ~p v ?p) == 1
(5.2)    R(p v ~p) + R?p == 1
(5.3)    Rp + R(~p v ?p) == 1
... etc

Using the relation (5.3) the definitions (1)...(4) can be transformed like
(with the prefix R used in definitions):

Tp    =df    (Rp = 1)
== (Rp = Rp + R(~p v ?p))
== (R(~p v ?p) = 0)
== U(~p v ?p)

...to:

(6)    Tp == U(~p v ?p)
(7)    Fp == E(~p v ?p)
(8)    Ep == F(~p v ?p)
(9)    Up == T(~p v ?p)

The relation between true and false remains unchanged. The relations between
T, F, E and U is:

Tp == ~Fp == ~E(~p v ?p) == U(~p v ?p)
Fp == ~Tp == ~U(~p v ?p) == E(~p v ?p)
Ep == ~Up == ~T(~p v ?p) == F(~p v ?p)
Up == ~Ep == ~F(~p v ?p) == T(~p v ?p)

Next we need to distinguish between necessary and temporary equality

In the relations above both '=' and '==' are used to express quantitative
equalities, '=' temporary equalities and '==' necessary equalities. If two
expressions are necessary equal they always has same truth value independent
of all circumstances and every arbitrary context or situation. If so, the
two expressions has both identical truthvalues and identical meaning.
Necessary equality '==' is stronger than '=' that only expresses temporary
equality of truth values, and stronger than '<=>' that only expresses
equivalens, i.e. 'true together' and 'false together'. Outside a bivalued
context equivalences are not sufficient to express that two expressions has
same meaning - we must prove they are necessary equal. From a necessary
equality we of course can conclude both temporary equality and equivalence.

Finally we need to enter modal concepts

Using the concept of necessary equality we can sharpen up (1)...(4) to
define modal concepts. Then 'p = 1' ( i.e. 'Rp = 1') is changed to 'p == 1'
we tells that 'p is necessary 100% true'. The negation of 'p = 1' is 'p <>
1' (i.e. in [0,1] the same as 'p < 1') will here correspond to 'p =/= 1', 'p
is not necessary 100% true'. L and M (like Lewis) are used as prefix for
'necessary true' and 'possible'.

(10)    Lp    =df    (Rp == 1)        ; 'p is necessary 100% true'
(11)    ~Lp    =df    (Rp =/= 1)      ; 'p is not necessary 100% true'
(12)    Mp    =df    (Rp =/= 0)       ; 'p is not necessary 0% true'
(13)    ~Mp    =df    (Rp == 0)       ; 'p is necessary 0% true'

In same way as above we can derive

(14)    Lp    ==    ~M(~p v ?p)
(15)    Mp    ==    ~L(~p v ?p)

We are now the whole turn around and back to a pure bivalued context. We
have still three alternatives, 'p', 'not-p' and 'unknown', but necessary
truth can not be mesured like the marbles in a urn as a portion of 100%.
There are no degrees of necessary truth - a statement is necessary true or
is not necessary true. There can not be other possibilities. However, by
value but not conceptually '=/= 1' is not the same as '== 0', so ~Lp will
never collaps into ~Mp like Fp does into Up in a contingent bivalued
context. As modal logic is pure bivalued it has been questioned what more it
can contribute to that not allready can be handled by standard logic. Modal
logic uses same rules of consequence as all logic but has quite other and
important properties than e.g. logic in a contingent context.

Intuitionism and constructivism has rejected reduction ad absurbum as an
unsure method of proof. Above is shown that from ~M~p we can not conclude
Lp. We need to show ~M(~p v ?p), i.e. both ~M~p ("it's not possible that
not-p") and ~M?p ("it's not possible that p is unknown") to conclude Lp.
Intuitionists are right in their critisism but hits the wrong property of
logic. Duoble negation reduction is a most basically foundation of all logic
and will still remain untouched. There are still no problems to reduce ~~p
to p, to toggle beween a meaning and its contrary meaning for every
negation. Also for every negation of truth we still can toggle between the
contraries Tp and Fp. As reductio ad absurbum proof takes place in modal
contexts we need to handle them by modal logic but for that we need a modal
logic strict constructive derived, like the logic refered above.

my best wishes
Åke Persson

```

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