In the Fregean Theory of Logic (the standard theory of logic in the
20th century),
“p implies q” has had been defined to mean “not-p or q”.

Here let P be “A lion is a mammal” and Q be “A lion and a whale are
both mammals”
It is clear that Q implies P but P does not imply Q.
Hence “P implies Q” is a falsity.
On the other hand, “not-P or Q” is a truth.
Therefore, it is a mistake to define “p implies q” to mean “not-p or
q”.

On Apr 29, 4:59 am, Eukie_M_SHIRAISHI <ms.eu...@gmail.com> wrote:

In the Fregean Theory of Logic (the standard theory of logic in the
20th century),
$B!H(Bp implies q$B!I(B has had been defined to mean $B!H(Bnot-p or q$B!I(B.

Here let P be $B!H(BA lion is a mammal$B!I(B and Q be $B!H(BA lion and a whale are
both mammals$B!I(B
It is clear that Q implies P but P does not imply Q.
Hence $B!H(BP implies Q$B!I(B is a falsity.
On the other hand, $B!H(Bnot-P or Q$B!I(B is a truth.
Therefore, it is a mistake to define $B!H(Bp implies q$B!I(B to mean $B!H(Bnot-p or
q$B!I(B.

Sorry, but I don't see why is $B!H(Bnot-P or Q$B!I(B true. I mean, what if a
lion and a whale are both not mammals?

Thank you for your following.

"not-P or Q�"means "A lion is not a mammal or a lion and a whale are
both mammals". Therefore, it is a truth-הסתר טקסט מצוטט-

-הר�ה טקסט מצוטט-

Thank you for your following.

"not-P or Q�"means "A lion is not a mammal or a lion and a whale are
both mammals". Therefore, it is a truth-הסתר טקסט מצוטט-

-הר�ה טקסט מצוטט-

Yes, but isn't "p->q" truth too?

I mean, if I remember correctly the truth value table of "implies", it
only gives F when p is F and p is T
Therefore, as here we have T->T, we have truth value T...

On 4$B7n(B30$BF|(B, $B8a8e(B6:44, ST <sara...@gmail.com> wrote:

Or have I misunderstood something?

In the Fregean Theory of Logic (the standard theory of logic in the
20th century),
“p implies q” has had been defined to mean “not-p or q”.

Here let P be “A lion is a mammal” and Q be “A lion and a whale are
both mammals”
It is clear that Q implies P but P does not imply Q.
Hence “P implies Q” is a falsity.

On the other hand, “not-P or Q” is a truth.
Therefore, it is a mistake to define “p implies q” to mean “not-p or
q”.

Since implication is one of the most fundamental concepts in logic,
all-out revision
is needed in the Fregean Theory of Logic.

Have a look at the following Web site:- <http://www.age.ne.jp/x/eurms/
PNG/Key/Honron-2.html#02-2

In the Fregean Theory of Logic (the standard theory of logic in the
20th century),
“p implies q” has had been defined to mean “not-p or q”.

Here let P be “A lion is a mammal” and Q be “A lion and a whale are
both mammals”

It is clear that Q implies P but P does not imply Q.
Hence “P implies Q” is a falsity.

On the other hand, “not-P or Q” is a truth.

Therefore, it is a mistake to define “p implies q” to mean “not-p or
q”.

Eukie_M_SHIRAISHI wrote:

In the Fregean Theory of Logic (the standard theory of logic in the
20th century),
$B!H(Bp implies q$B!I(B has had been defined to mean $B!H(Bnot-p or q$B!I(B.

Here let P be $B!H(BA lion is a mammal$B!I(B and Q be $B!H(BA lion and a whale are
both mammals$B!I(B
It is clear that Q implies P but P does not imply Q.
Hence $B!H(BP implies Q$B!I(B is a falsity.

Not so. Since Q is true, P implies Q.

"A lion and a whale are both mammalds " does imply "A lion is a
mammal"
but "A lion is a mammal" does NOT imply "A lion and a whale are both
mammalds "

"A lion and a whale are both mammalds " does imply "A lion is a
mammal"
but "A lion is a mammal" does NOT imply "A lion and a whale are both
mammalds "

There is a difference between the boolean "implies" and "being a
logical
consequence".