Re: A missing definition in "Gödel's Proof" by Nagel & Newman (open letter)

translogi <wilemien@xxxxxxxxxxxxxx> writes:

On Nov 2, 7:25 pm, Alan Smaill <sma...@xxxxxxxxxxxxxxxx> wrote:
translogi <wilem...@xxxxxxxxxxxxxx> writes:
On Nov 1, 11:10 pm, Peter_Smith <ps...@xxxxxxxxx> wrote:
It was asked:

Is a truth-table for that statement that shows it to be true in
all combinations of T and F for p and q considered to be a valid
proof of its truth?

If that means "a valid proof in deductive system S", the answer for
most systems S is, of course, trivially "no", since a proof-in-system-
S is typically defined otherwise (in terms of e.g. a sequence of axiom-
instances, or applications of rules of inference to previous items in
the sequence). That's trivial.

What I was commenting on was the unqualified claim that

Truth tables do not prove anything.

That is plainly over-stating the case. For of course, giving a truth-
table can prove that a wff is a tautology, and hence true. [If I set
students an exercise "prove ¬¬(P v ¬¬¬P) is a tautology" and they do a
correct truth-table, they get full marks! :-) ]

Do you also teach intuitionistic logic? (just joking it isn't a
tautology in that system that is why, but i know that you know that.)

isn't it?

In intuitionistic logic (A v ~A) is not a theorem
neither is ~~A -> A
neither is ((A -> B) & (~A -> B)) -> B
and there is also a fourth from that also is not a theorem.

(they are all interchangable)



A -> ~~A is a theorem
A -> (~A -> B ) is also a theorem (but not in minimal logic)

right, but the formula in question is not (A v ~A), it is
(equivalent to) ~~(A v ~A);
and that is intuitionistically provable.

--
Alan Smaill
.

Relevant Pages