@qaanol, on Jan 7 2009, 11:30 AM, said in DeMorgan's Rule:

Please do not confuse the logical "then" with the concept of "causes". The fallacy described amounts to "Correlation implies causation." But the logical argument presented amounts to "Correlation implies tautology." Since tautology is always true, the argument is valid.

I acknowledge that there may be confusion about the term "causation." The statement:

A implies B

only means that

It does not imply that in natural terms, A is the cause per se of B, according to the natural concept.

The fallacy of your argument lies in the order in which you evaluated your truth table.

Note: None of my following argument would be true if p is always true or if q is always true, because if p is always true, then q -> p. However, neither of those statements can be derived from the original premise, which is: for **some** x, p(x) and q(x)

```
| p | q | p^q | p->q| q->p| (p->q)v(q->p)
| T | T | T | T | T | T
| T | F | F | F | T | T
| F | T | F | T | F | T
| F | F | F | T | T | T
```

The last column is not valid.

The conclusion of the statement requires that

p -> q for **all** x.

or

q -> p for **all** x.

because **the domains of p and q** , as you defined them, are all x.

Some definitions:

X: the domain of all possible values of x.

X1: values of x such that p and q

X2: values of x such that p and not q

X3: values of x such that (not p) and q

X4: values of x such that (not p) and (not q)

X1..4 are mutually exclusive, and X is the union of them.

Now, a final row showing the logical AND of each individual possibility, which is: is this statement always true?

```
x in| p | q | p^q | p->q| q->p| (p->q)v(q->p)
X1 | T | T | T | T | T | T
X2 | T | F | F | F | T | T
X3 | F | T | F | T | F | T
X4 | F | F | F | T | T | T
====+=====+=====+=====+=====+=====+=====
X | F | F | F | F | F | F
```

p is not true for all x in X, because it is not true in X3 and X4.

true likewise.

p ^ q is not true for all x in X, because it is only true in X1.

p->q is not true for all x in X, because it is not true for x in X2.

q->p likewise.

@guy, on Jan 7 2009, 12:08 PM, said in DeMorgan's Rule:

The argument is logically valid, yes, but it is a fallacy when used in the real world, eg. a court of law.

It is invalid in court because it is invalid in logic, Q.E.D.

(Now I feel odd, not using the mathematical symbols, after making a deal about that. But there are a **lot** )

(That could probably be reordered to follow a better pattern, but I'm making it more of a reply, depending on previous material).