The general form of THE SECURABILITY FORMULA S{-}-- involving FAC ("fallacy of asserting the consequent -- IT IS:

S{FAC} = S{((&iTi -> {&jAj)& (&jAi)) -> &iTi} = 1 - 1/2j + 1/2i+j [1]

For i = j = 1, FAC is:

	((T -> A) & A) -> T [1A]

There is another "fallacy" derived from the conditional T -> A, it is FC ("fallacy of the contrary"):

		A -> T. [1B]

The INDICATOR ("TRUTH") TABLES for both are:

T	A	FAC	FC	FAC<->FC
0	0	1	1	1
0	1	0	0	1
1	0	1	1	1
1	1	1	1	1

Clearly, the third and fourth TABLES are THE SAME. This means that FC and FAC ARE LOGICALLY EQUIVALENT: FAC = FC. And that is also what the fifth table says (FAC IF, AND ONLY IF, FC). This latter STATEMENT is a TAUTOLOGY. Let's see what that means.

THE RULE OF SUBSTITUTION ALLOWS US TO EXTEND THIS EQUIVALENCE FOR ALL i, j. We see that FAC = FC, an EQUIVALENCE, wherein i = j = 1. We extend FAC to have i = 1, 2, ..., n TIPOFF's and j = 1, 2, ..., n ANOMALIES. Now, for every T_i in the EXTENDED FAC we insert the same in EXTENDING FC; and for every A_j in the EXTENDED FAC we insert the same in EXTENDING FC. Hence, they remain EQUIVALENT in EXTENSION, that is, for i = 1,2,...,n and j = 1,2,...,n. What's the advantage of this?

FAC has 2i + 2j + 1 CONJUNCTIONS and 2 CONDITIONALS; FC has only i + j CONJUNCTIONS and 1 CONDITIONAL -- much simpler to deal with.

So, the problem is, now, to prove:

	S{FC} =S{(&j
Aj) Þ(&iHi)} = 1 - 1/2j
+ 1/2i+j  [2]

In this form, we ask, "When is the Indicator of the Predecessor equal to 1 when the indicator of the Consequent equals 0?" For, in such case, the Indicator of the CONDITIONAL is 0; otherwise, it is 1. (For the case of i = 2, j = 3, You can confirm the statements that follow.

The GENERAL INDICATOR TABLE is written for FAC, with the i TIPOFF's on the LEFT SIDE of the TABLE, and the j ANOMALIES on the RIGHT. (The rows of the TABLE are equivalent to the BINARY NOTATION of the decimal numbers for 1, 2, ..., 2^i+j.) But our ANALYSIS proceeds as for FC, considering first THE CONJUNCTION of the j ANOMALIES first, and THE CONJUNCTIONS OF THE i TIPOFF's.

Now, A CONJUNCTION OF STATEMENTS IS "TRUE" (INDICATOR 1) IF, AND ONLY IF, ALL CONJUNTANDS ARE "TRUE" (INDICATORS ALL 1), OTHERWISE IT IS "FALSE" (NOT ALL INDICATORS 1).

• FOR THE ANOMALIES (forming the PRECEDENT of our CONDITIONAL), the CONJUNCTION INDICATOR OF 1 OCCURS ONLY EVERY jth POSITION, hence, occurs 2 ⁱ TIMES, all other entries being 0.

• FOR THE CONJUNCTION OF THE TIPOFF's (CONSEQUENT of our CONDITIONAL), ONLY THE LAST j INDICATORS ARE 1, THE REST BEING 0. This means that ALL EXCEPT ONE OF THE 2ⁱ INDICATOR of 1 for the PRECEDENT matches a 0 INDICATOR for the CONSEQUENT. Therefore, THE CONDITIONAL OF FC has INDICATOR 0 in EXACTLY 2 ⁱ - 1 instances (rows of the TABLE), the REST being 1's.

• HOWEVER, WE ARE INTERESTED IN THE 1's (RATHER THAN 0'S) OF FC (IN COMPUTING THE SECURABILITY MEASURE), AND THIS IS THE COMPLEMENT OF THE ABOVE RESULT. SO WE SUBTRACT THE ABOVE RESULT FROM THE TOTAL (2^i+j):

  
  	2i+j - (2i - 1) = 2i+j - 2i + 1 [3]

• But we wish this to be "normalized", that is, formulated as a FRACTION OF THE TOTAL, hence, we DIVIDE [3] BY 2^i+j:

  
  	(2i+j - 2i + 1)/2i+j = 1 - 1/2j + 1/2i+j [4]

  which AGREES WITH [1]. QED.

  Please note that the FORMULA relates to two DISTINCT (but EQUIVALENT) LOGICAL FORMS. I discovered this for myself and have seen it nowhere in "the literature". Hence, it is convenient to introduce a mediary label, by way of defining THE OCKAM FUNCTION, O(i, j), involving i TIPOFF's and j ANOMALIES.

  Then, we may restate our result:

    S(O(i,j)) = 1 - 1/2j + 1/i+j [1C].
   

  Math students may notice that my SECURABILITY FORMULA has the FORM of the ALGORITHM of INCLUSION AND EXCLUSION.