Question

\[(f\Rightarrow t)\wedge (f\Rightarrow f)\]

Answer 1

Yes.

Answer 2

can you explain why , please

Answer 3

Oh, I was being a jerk. I can't actually read the question, and thought it wasn't a yes/no question.

My bad.

Answer 4

well the statement is true,

Answer 5

I'm sorry, what's the middle symbol?

Answer 6

conjunction, kinda like AND

Answer 7

That's what I thought. But then the statement has to truth value, does it? F maps to T, and then F maps to itself which is the... reflexive property.

Answer 8

So is it actually saying something, or am I misunderstanding.

Answer 9

im not really used to the \(\Rightarrow\) symbol
in this question it means a hypothetical proposition

Answer 10

The best way I can describe what I'm thinking is that you only stated a presupposition, but asked no questions.

F maps to T, and F maps to F. The first part is the presupposition. The second part is identity.

I don't think you can infer anything else from this.

Answer 11

what in gods name is that?? yea...i'm saving myself before it's to late.

Answer 12

GOODNIGHT EVERBODY ;D

Answer 13

It's a statement. I don't actually see a question.

Answer 14

a statement can be true or false

Answer 15

Okay, this statement is non-contradictory. So probably true as a result. You define a function \(f\) to be mapped to \(t\). Then you add the condition that \(f\) also maps to \(f\). Unless I'm very well mistaken, the property of identity is an axiom, so neither condition falsifies the other, and you have a statement that is ostensibly true.

It's like saying, "If \(p\) then \(q\), and if \(p\) then \(p\)." Sure, that's true.

Because of this straightforwardness, I think I'm misinterpreting some symbol somewhere.

Answer 16

you can make a truth table if you like

Answer 17

\[\begin{array}{ccc}\ \psi&\phi&\psi\Rightarrow\phi\\ \text{____}&\text{____}&\text{____}\\t&t&t\\ t&f&f\\ f&t&t\\f&f&t \end{array}\]


how can it map to two places at once?

Answer 18

the statement \((p\to q)\land (p\to p)\) is not always true

Answer 19

Ignore me, then, I'm misunderstanding something!

Answer 20

\(p\to p\) is a tautology, but the conjunction is not always true

Answer 21

that is to say \((p\to q)\land (p\to p)\) is identical to \(p\to q\)

Answer 22

The tautology doesn't modify the p to q truth value at all, so if we presuppose that p to q, isn't it tautological to say that p to q is true?

Answer 23

t for true
f for false

Answer 24

maybe i do not understand the question
to say something like "this is true" means to me that it is always true, i.e. that it is a tautology. like for example \(p\lor \lnot p\)

Answer 25

thats right

Answer 26

Maybe the second tautology isn't a tautology. The premise f is false, maybe, is what it's saying. Since you have the premise f being both true and false (f to t, and f to f-alse) you have a contradiction.

Answer 27

but it is entirely possible i did not understand the question, because there is not a question here, just a statement \((f\to t)\land (f\to f)\)

Answer 28

My original issue with it. I'm thinking that not all the f's are the same.

Answer 29

i am not really sure what is being asked, but it is not the case that the above statement is TRUE in the sense that i can be replaced in a truth table by \(T\)

Answer 30

oh