# Thread: How is it true? p <=> p v p<=> (~p=>p).

1. ## How is it true? p <=> p v p<=> (~p=>p).

Hello everybody.

We have p <=> p v p<=> (~p=>p).
I don't understand it. For example assume p: (2+2=4), so by this logical statements we have:

(2+2=4)=> {if (2+2 is not equal to 4) then (2+2=4)}.

and

{if (2+2 is not equal to 4) then (2+2=4)}=> (2+2=4)

Thanks.

2. Originally Posted by Researcher

p <=> p v p<=>(~p=>p).
The above should be ~(~p=>p). It's the same as saying, "It is false that {If 2+2 $\not=$4, then 2+2=4.}"

If p is true, then p v p is true. In logical expression $p \Rightarrow p \vee p$

The conjunction is true if at least one is true.

If the p v p is true, then p is true. In logical expression $p \vee p \Rightarrow p$

$(p \Rightarrow p \vee p) \vee (p \vee p \Rightarrow p)$ is equivalent to $p \Leftrightarrow p \vee p$.

Now $p \vee p \Leftrightarrow \sim \sim p \vee \sim \sim p$.

De Morgan's Law says $\sim \sim p \vee \sim \sim p\Leftrightarrow \sim (\sim p \wedge \sim p)$

$\sim (\sim p \wedge \sim p) \Leftrightarrow \sim (\sim p \Rightarrow p)$

---
The conjunction $p \vee p$ is like you are being offered to have a potato or a potato. There is only one choice, potato.

It's different if the conjuction is $p \vee q$, which is like being offered rice or potato. If it's rice then it's not potato. So either way, you get at least one.

$p \Rightarrow p \vee p$ is like the same as saying, "If it's a potato, then you get potato or potato."

Here is the De Morgan's Law $\sim(P \vee Q) \equiv \sim P \wedge Q$.

Here is the negation of implication: $\sim (P \Rightarrow Q) \equiv P \wedge \sim Q$

3. Originally Posted by novice
The above should be ~(~p=>p). It's the same as saying, "It is false that {If 2+2 $\not=$4, then 2+2=4.}"

If p is true, then p v p is true. In logical expression $p \Rightarrow p \vee p$

The conjunction is true if at least one is true.

If the p v p is true, then p is true. In logical expression $p \vee p \Rightarrow p$

$(p \Rightarrow p \vee p) \vee (p \vee p \Rightarrow p)$ is equivalent to $p \Leftrightarrow p \vee p$.

Now $p \vee p \Leftrightarrow \sim \sim p \vee \sim \sim p$.

De Morgan's Law says $\sim \sim p \vee \sim \sim p\Leftrightarrow \sim (\sim p \wedge \sim p)$

$\sim (\sim p \wedge \sim p) \Leftrightarrow \sim (\sim p \Rightarrow p)$

---
The conjunction $p \vee p$ is like you are being offered to have a potato or a potato. There is only one choice, potato.

It's different if the conjuction is $p \vee q$, which is like being offered rice or potato. If it's rice then it's not potato. So either way, you get at least one.

$p \Rightarrow p \vee p$ is like the same as saying, "If it's a potato, then you get potato or potato."

Here is the De Morgan's Law $\sim(P \vee Q) \equiv \sim P \wedge Q$.

Here is the negation of implication: $\sim (P \Rightarrow Q) \equiv P \wedge \sim Q$
Thanks.

We have; $p \Leftrightarrow p \vee p$
and
$(p \vee q) \Leftrightarrow (\sim p \Rightarrow q)$ (by the law of contraposition)

So if choosing p instead of q, we have;
$p \Leftrightarrow p \vee p \Leftrightarrow (\sim p \Rightarrow q)$
So what's wrong now?

4. Originally Posted by Researcher
Thanks.

We have; $p \Leftrightarrow p \vee p$
and
$(p \vee q) \Leftrightarrow (\sim p \Rightarrow q)$ (by the law of contraposition) <---this is wrong.

So if choosing p instead of q, we have;
$p \Leftrightarrow p \vee p \Leftrightarrow (\sim p \Rightarrow q)$
So what's wrong now?
A contrapositive statement is the converse of an implication.

If the implication is $p \Rightarrow q$, the contrapositive is $q \Rightarrow p$

In other words, only an implication would have a contrapositive statement.

5. Originally Posted by novice
A contrapositive statement is the converse of an implication.
If the implication is $p \Rightarrow q$, the contrapositive is $q \Rightarrow p$
In other words, only an implication would have a contrapositive statement.
That is wrong.
The converse of $p \Rightarrow q$ is $q \Rightarrow p$.
The contrapositive of $p \Rightarrow q$ is $
\neg q \Rightarrow \neg p$
.

It is the case that $\left[ {p \Rightarrow q} \right] \equiv \left[ {\neg p \vee q} \right]\,\& \,\neg \left[ {p \Rightarrow q} \right] \equiv \left[ {p \wedge \neg q} \right]$

6. Originally Posted by Plato
That is wrong.
The converse of $p \Rightarrow q$ is $q \Rightarrow p$.
The contrapositive of $p \Rightarrow q$ is $
\neg q \Rightarrow \neg p$
.

It is the case that $\left[ {p \Rightarrow q} \right] \equiv \left[ {\neg p \vee q} \right]\,\& \,\neg \left[ {p \Rightarrow q} \right] \equiv \left[ {p \wedge \neg q} \right]$
Yes, I have been very careless.

The converse is $q \Rightarrow p$, but the contrapositive is $\sim q \Rightarrow \sim p$.

Thank for pointing to it. I have to weed out my bad habit.

7. Originally Posted by Researcher
We have p <=> p v p<=> (~p=>p).
I hope this formula is not parsed as $(p\Leftrightarrow p)\lor(p\Leftrightarrow(\sim p\Rightarrow p))$, because otherwise I have to think long and hard to decide whether this is true!

Originally Posted by novice
The above should be ~(~p=>p).
No, it is correct as stated in the original post. For all $p$ and $q$ we have $(p\Rightarrow q)\Leftrightarrow(\sim p\lor q)$, so $\sim p\Rightarrow p$ is equivalent to $\sim\sim p\lor p$, or $p\lor p$. Maybe the mistake was here:
(outside negation has to be removed from either the left- or the right-hand side).

8. Thanks friends.
So i am right!

9. Originally Posted by emakarov
...outside negation has to be removed from either the left- or the right-hand side).
Can't picture it in my head.

$P \wedge \sim Q \equiv \sim (P \Rightarrow Q)$

Now $\sim P \wedge \sim Q \equiv \sim (\sim P \Rightarrow Q)$

Then $\sim (\sim P \wedge \sim Q) \equiv \sim \sim (\sim P \Rightarrow Q)$

Substituting $P$ for $Q$, we obtain

$\sim (\sim P \wedge \sim P) \equiv (\sim P \Rightarrow P)$

Let P be the statement: Pigs fly.
$\sim P \Rightarrow P$: If pigs don't fly, then pigs fly. Oh, lovely.

I know, I know, $F \Rightarrow T$ is $T$. That's why in math, it's called Trivial Proof, a useless proof.

$F \Rightarrow F$ is true; it's called Vacuous Proof, another useless proof.

10. So, don't you think it is a dubious logical expression?!
I'm really getting crazy with this!!

11. Originally Posted by Researcher
So, don't you think it is a dubious logical expression?!
I'm really getting crazy with this!!
No friend, they are no dubious logical. They have survived the test since ~350 B.C.