Find truth value of propositions

[evinda]

Hello! (Wave)

We suppose that the propositions $p,q$ are propositions such that the proposition $p \to q$ is false.
Find the truth values for each of the following propositions:

1. $\sim q \to p$
2. $p \land q$
3. $q \to p$

I have thought the following:

Since the proposition $p \to q$ is false, either $p$ is true and $q$ is false, either $q$ is true and $p$ is false.

Thus, we have the following truth table:

$\begin{matrix}
p & q & \sim q & \sim q \to p & p \land q & q \to p\\
0 & 1 & 0 & 1 & 0 & 0\\
1 & 0 & 1 & 1 & 0& 0
\end{matrix}.$

So, $\sim q \to p$ is at each case $1$ since $\sim q$ and $p$ have the same values.

As for $p \land q$ it is always $0$ since $p$ and $q$ have different values.

$q \to p$ is for the same reason $0$.

Is everything right? Or am I somewhere wrong?

 

[HallsofIvy]

I would note immediately that "∼q→p" is equivalent to "p→q" so it is false.
I fact, of the four combinations of "true" and "false" for p and q the only case in which
p→q is false is p true and q false. In that case p∧q is false but q→p is true.

 

[I like Serena]

Since the proposition $p \to q$ is false, either $p$ is true and $q$ is false, either $q$ is true and $p$ is false.

Hey evinda!

As HallsofIvy already pointed out, I'm afraid this is not correct. (Worried)

 

[evinda]

I would note immediately that "∼q→p" is equivalent to "p→q" so it is false.
I fact, of the four combinations of "true" and "false" for p and q the only case in which
p→q is false is p true and q false.

I haven't understood why $p \to q$ is only false when p is true and q is false. Isn't it also false when p is false and q is true?

In that case p∧q is false but q→p is true.

Why is in this case $q \to p$ true?

 

[evinda]

Hey evinda!

As HallsofIvy already pointed out, I'm afraid this is not correct. (Worried)

I haven't understood why "∼q→p" is equivalent to "p→q" .. (Sadface)

 

[I like Serena]

The implication $p\to q$ is false iff p is true and q is false.

However, it is not equivalent to $\lnot q\to p$ being false.
Instead it is equivalent to $\lnot q\to\lnot p$ being false.

 

[evinda]

The implication $p\to q$ is false iff p is true and q is false.

Could you explain to me why this happens? :unsure::unsure::unsure:

However, it is not equivalent to $\lnot q\to p$ being false.
Instead it is equivalent to $\lnot q\to\lnot p$ being false.

In general it holds that $p \to q \iff \lnot q \to \lnot q$, right? (Thinking)

 

[I like Serena]

Could you explain to me why this happens?

The implication $p\to q$ means that if $p$ is true that $q$ must also be true.
It follows that the implication must be false if $p$ is true but $q$ is false, doesn't it?

That leaves the other 2 cases where $p$ is false.
They have been defined such that $p\to q$ is true if $p$ is false.
That's all there is to it. :geek:

In general it holds that $p \to q \iff \lnot q \to \lnot p$, right?

Yep. (Nod)

 

[evinda]

The implication $p\to q$ means that if $p$ is true that $q$ must also be true.
It follows that the implication must be false if $p$ is true but $q$ is false, doesn't it?

So if we have a proposition , say, $x \to y$, it is considered that $x$ is true, right?
And so if the proposition is false, given that $x$ is true, it must hold that $y$ is false, right?

That leaves the other 2 cases where $p$ is false.
They have been defined such that $p\to q$ is true if $p$ is false.
That's all there is to it. :geek:

The case that $p$ is false doesn't need to be checked, since we assume that $p$ is true, right? Or have I understood it wrong? (Thinking)

 

[I like Serena]

So if we have a proposition , say, $x \to y$, it is considered that $x$ is true, right?
And so if the proposition is false, given that $x$ is true, it must hold that $y$ is false, right?

Yep. (Nod)

The case that $p$ is false doesn't need to be checked, since we assume that $p$ is true, right? Or have I understood it wrong?

Generally we also need to check cases where $p$ is false.
However, for this particular problem we must have that $p$ is true and $q$ is false, since otherwise $p\to q$ wouldn't be false.

 

[evinda]

Yep. (Nod)

How can this be shown that $p \to q \iff \lnot q \to \lnot p$ ? (Thinking)

 

[evinda]

So, in this case we have that $p \to q$ is false. That means that $p$ is true and since the proposition is false we get that $q$ is false.
We want to find the truth value of $\lnot p \to q$.

$p \to q$ is false iff $\lnot p \to q$ is true,since $ \lnot p$ is false and so is $q$. Is this right and sufficient? (Thinking)

 

[I like Serena]

How can this be shown that $p \to q \iff \lnot q \to \lnot p$ ?

One way to show it, is to consider that the left hand side is false if and only if $p=\text{true}$ and $q=\text{false}$.
If we have indeed that $p=\text{true}$ and $q=\text{false}$, then $\lnot q=\text{true}$ and $\lnot p=\text{false}$, isn't it?
So $\lnot q \to \lnot p$ is false if and only if $p=\text{true}$ and $q=\text{false}$.

Alternatively we can make a truth table for both expressions and see that those truth tables are identical.

This is not relevant for the problem at hand though is it? (Wondering)

 

[evinda]

One way to show it, is to consider that the left hand side is false if and only if $p=\text{true}$ and $q=\text{false}$.
If we have indeed that $p=\text{true}$ and $q=\text{false}$, then $\lnot q=\text{true}$ and $\lnot p=\text{false}$, isn't it?
So $\lnot q \to \lnot p$ is false if and only if $p=\text{true}$ and $q=\text{false}$.

Alternatively we can make a truth table for both expressions and see that those truth tables are identical.

Ok... (Nerd)

This is not relevant for the problem at hand though is it? (Wondering)

Is the explanation I wrote above sufficient for the specific problem?

 

[I like Serena]

Is the explanation I wrote above sufficient for the specific problem?

You have distinguished the cases $p=1,q=0$ and $p=0,q=1$.
But that second case cannot occur since $0\to 1$ is true.
So we should only look at the first case: $p=1, q=0$.

It means in particular that:
$$(q\to p) = (0\to 1) = 1$$
which is not what you had. (Worried)

 

[evinda]

You have distinguished the cases $p=1,q=0$ and $p=0,q=1$.
But that second case cannot occur since $0\to 1$ is true.
So we should only look at the first case: $p=1, q=0$.

It means in particular that:
$$(q\to p) = (0\to 1) = 1$$
which is not what you had. (Worried)

We have the following truth table:

\begin{equation*}
\begin{array}{c|c|c|c|c}
p & q & p\to q & \lnot q & \lnot q \to p & p\land q & q \to p\\
\hline
1 & 1 & 1 & 0 & 1 & 1& 1\\
1 & 0 & 0 & 1 & 1& 0& 1\\
0 & 0 & 1 & 1 & 0& 0& 1\\
0 & 1 & 1 & 0 & 1& 0&0
\end{array}
\end{equation*}

Since $p \to q$ is false, we are interested in the case when $p=1$ and $q=0$. Thus $\lnot q \to p$ is true, $p \land q$ is false and $q \to p$ is true. Right?

So we didn't need to complete the whole truth table sincechecking the truth values of $p \to q$ at the beginning, we see that it can hold just for $p=1$ and $q=0$, right? :unsure:

 

[I like Serena]

Since $p \to q$ is false, we are interested in the case when $p=1$ and $q=0$. Thus $\lnot q \to p$ is true, $p \land q$ is false and $q \to p$ is true. Right?

So we didn't need to complete the whole truth table sincechecking the truth values of $p \to q$ at the beginning, we see that it can hold just for $p=1$ and $q=0$, right?

Yep. (Nod)

 

[evinda]

Yep. (Nod)

Nice, thank you!