Please recall that we've seen the simple form of FAC, peoratively labelled "fallacy of asserting the consequence", which (unlike MP and other tautologies) gives FAC the potential of referring to Reality.

We labeled its STATEMENTS as "T, E" -- "T" for "TIPOFF", "A" for "ANOMALY". So it runs: ((T -> E) & E) -> T. That is,

• (T -> E): "Tipoff T implies Anomaly A".

• (...)& E: "Exception A is OBSERVED".

• ((...) & E) -> T: "This EVALUATES T as (possibly) CORRECT".

It has the TABLE:

T	A	T -> A	(T -> A) & A	((T -> A) & A) -> T
0	0	1	0	1
0	1	1	1	0
1	0	0	0	1
1	1	1	1	1

Please notice that the final Table (Column) has 3 1's (for "True") and a single 0 (for "False"). One possibility out of four for "going wrong". Let's "quantify" that by defining a MEASURE, which will lead to our "big" measure: SECURABILITY.

DEFINITION: The BASE of a TABLE equals the NUMBER of SIGNS it contains.

Above, BASE = 4.

(As some of you may know, BASE always has the FORM 2^t, where t is THE NUMBER OF TIPOFF's involved -- since each is BIVALENT -- "CORRECT" or "INCORRECT" -- "1" or "0". This follows as a Theorem from the BASE DEFINITION and the TABLE DEFINITION.)

DEFINITION: The TALLY of a TABLE equals the NUMBER of ONES it contains.

DEFINITION: The SECURABILITY of a LOGICAL FORM, S(Form), is the RATIO (in its FINAL TABLE) of TALLY to the BASE: S(Form) = TALLY/BASE.

Above, S(FAC) = 3/4.

We also looked at the case wherein THE SAME TIPOFF YIELDED ANOTHER ANOMALY, AND IT WAS OBSERVED. That is, one TIPOFF, T, two ANOMALIES, A₁ and A₂. This took the FORM, ((T -> (A₁ & A₂) & A₁ & A₁)) -> T.

Here are the TABLES for this:

T	A1	A2	(A1&A2	T->(A1&A2)	((T->(A1&A2))&(A1&A2)	(((T->(A1&A2))&(A1&A2))->T
0	0	0	0	1	0	1
0	0	1	0	1	0	1
0	1	0	0	1	0	1
0	1	1	1	1	1	0
1	0	0	0	0	0	1
1	0	1	0	1	0	1
1	1	0	0	1	0	1
1	1	1	1	1	1	1

The BASE for this is 8. Again, a single "0". So its SECURABILITY is 7/8 > 3/4.

Hey, looky! Because T yielded one more OBSERVED ANOMALY than previously, the SECURABILITY HAS INCREASED!

What would happen if, given the same single TIPOFF, T, another OBSERVED ANOMALY was derived? BASE would DOUBLE, from 8 to 16, with still a single "0", so its SECURABILITY WOULD BECOME 15/16 -- ANOTHER INCREASE.

But some cases require increases in the tipoffs, so we need a FORMULA for the GENERAL CASE, NAMELY, (((&_iT_i) - (&_jA_j)) & (&_jA_j))) -> &_iT_i, where i = 1,2,...; j = 1,2,...

That is, a number (i) of tipoffs IMPLY another number (j) of anomalies, which are OBSERVED; so we CLAIM "TRUTHNESS" OF THE TIPOFF SUM.

We need a FORMULA on the numbers i, j in this GENERAL FORM. For reason given in a following file, we'll call this GENERAL FORM, "The Ockam Function", labeling it as "O(i,j)", a function of S(O(i,j)) = 1 - 1/2^j + 1/2^i+j.

Try it on the simplest case, namely, O(1,1): S(O(1, 1)) = 1 - 1/2¹ + 1/2¹⁺¹ = 1 - 1/2 + 1/4 = 3/4. YEAH!

Try it on that second case (above), namely, O(1, 2): S(O(1,2)) = 1 - 1/2² + 1/2¹⁺² = 1 - 1/4 + 1/8 = 7/8. UHUH!

The great thing about O(i,j), is what happens in the case O(1,j) wherein j INCREASES -- that is, THE SAME TIPOFF YIELDS MORE AND MORE OBSERVED ANOMALIES.

S(O(1,j)) = 1 - 1/2^j + 1/2^{1 + j} = (2^1+j - 2 + 1)/2^{1 + j} = (2^1+j - 1)/2^1+j = 1 - 1/2^j+1.

In that LAST FRACTION:

1. The numerator remains CONSTANT at 1;
2. The denominator INCREASES with j;
3. So this fraction "grows smaller and smaller";
4. For some large value of j, say, j = 100, it is "subtracting practially nothing from the 1":
5. So, with increased success, O(1,j) -- GENERALIZED FAC -- has SECURABILITY APPROACHING 1! It's almost as good as the TAUTOLOGY, MP!

(Students of "Calculus" will note resemblance to many "limit processes".)

In fact, we find the above result as a theorem along with other BONUSES (theorems) of this FORMULA.