[FOM] Criteria for a correct assignment of T, F, and Para

Sandy Hodges SandyHodges at attbi.com
Mon Sep 30 21:37:57 EDT 2002

An assignment of the truth values: True, False, and Paradoxical, to a
finite net of sentence tokens is *correct* if it satisfies these four

1.   AUDIT TEST.     Here's a net of tokens as an example:

1.  Tr(3) & Fa(4)
2.  Para(1) & Fa(3)
3.  Tr(1) <=> Para(2)
4.  Fa(3)

Given an assignment to these tokens, for example <Para,T,F,F>, we can
calculate a new assignment vector.   Token 1 says 3 is true and 4 is
false, while the assignment says 3 is false and 4 is false, so token 1
has the calculated value: false.    Doing the calculation for all tokens
gives a new vector <F,T,T,T>.   If the original vector of truth values
is called G, then the calculated one is E(G).  An assignment G passes
the audit test if every token that gets T or F in G, gets the same truth
value in E(G).

Note that E(A), for any A, is a vector of T's and F's.    If G calls a
token Para, then E(G) will call that token either T or F.   But to pass
audit, only those tokens called T or F by G must remain the same in

We'll say a vector H *improves* on vector G if the set of tokens which H
calls Para is a proper subset of those which G calls Para, and if H
agrees with G about all the tokens G calls T or F.

As we look for vectors which pass the first three tests, we start
scanning those vectors that call no tokens Para, then those that call
one token Para, and so on.   Thus, when we are looking at a vector G
which calls N tokens Para, we already know which vectors pass the three
tests, for those vectors which call Para fewer than N tokens.   In
particular, of the vectors which improve on G, we know which vectors
pass the three tests.

2.  SPIN TEST.   For a vector G which passes the audit test, consider
the set M(G), containing all vectors which improve on G and which pass
the first three tests.    If M(G) contains nothing, G passes the spin
test.   If M(G) contains exactly one vector, G fails the spin test.
If M(G) contains two or more vectors, then if G does pass spin (and also
passes tests 1 and 3) then G will be called the roof of the set M(G).
The rule for whether G does pass when M(G) contains 2 or more vectors,
is that G passes unless there is a vector H in M(G), which is already
the roof of (M(G) - {H}).

3.  UNIQUENESS TEST.    If G passes the audit and spin tests, and if
there is another vector H, which calls the same tokens Para that G calls
Para, and if H passes the audit and spin tests, then both G and H fail
the uniqueness test.   If there is no such H, G passes.

As we test vectors which call N tokens Para, we can scan first those
that call Para a particular set of N tokens, then all those that call
Para another set of N tokens, and so on.    As we scan each vector we
can try the audit and spin tests.   After we have finished with the the
vectors that call a particular set of tokens Para, we can apply the
uniqueness test.    A vector will only pass if it was the only one to
pass the first two tests for this set.

4.  ALTERNATIVES TEST.   A vector G passes this test if it is not in a
set roofed by another vector that passes the first three tests.
Proof that there is one and only one correct vector.

Lemma: there is at least one vector which passes the first 3 tests.
The vector W which calls all tokens Para must pass the audit test, and
if it passes the spin test must pass the uniqueness test.   So W can
only fail if it fails the spin test, and W doesn't fail the spin test if
M(W) is empty.    So either W passes all 3 tests or M(W) is not empty,
and any vector in M(W) passes the first 3 tests.

Lemma: at least one unroofed vector passes the first 3 tests.
Any operation that roofs vectors, leaves the new roof unroofed.   So
there is no way to roof the last unroofed vector and leave none

So there is at least one unroofed vector passing tests 1 - 3.   Suppose
there are two or more, and choose two of them, A and B.  Consider the
vector W which calls all tokens Para.   Both A and B are in M(W).    If
W passes all 3 tests, then it roofs M(W) and by assumption A and B are
not in a roofed set.    So W must fail.   W can't fail audit, and if it
passes the spin test can't fail uniqueness.   So W must fail spin.   But
since M(W) contains at least two vectors, W can only fail if there is a
vector H which is the roof of (M(W) - {H}).    So one or both of A and B
are roofed by this H, contrary to assumption.   So there are not two or
more unroofed vectors.   Therefore, of the vectors that pass tests 1 -
3, one and only one also passes test 4, QED.
The important thing about this algorithm, is that it does not rely on
finding which tokens refer to which other tokens.    Thus it
"automatically" detects idle reference.

------- -- ---- - --- -- --------- -----
Sandy Hodges / Alameda,  California,   USA
mail to SandyHodges at attbi.com will reach me.

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