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Find truth value of propositions

evinda

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Apr 13, 2013
3,704
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

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MHB Math Helper
Jan 29, 2012
1,151
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.
 

Klaas van Aarsen

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Mar 5, 2012
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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

Well-known member
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Apr 13, 2013
3,704
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? :oops:
 

evinda

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Apr 13, 2013
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Klaas van Aarsen

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Mar 5, 2012
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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

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Apr 13, 2013
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Klaas van Aarsen

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Mar 5, 2012
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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

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Apr 13, 2013
3,704
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)
 

Klaas van Aarsen

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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

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Apr 13, 2013
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evinda

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Apr 13, 2013
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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)
 

Klaas van Aarsen

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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

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Apr 13, 2013
3,704
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? 🤔
 

Klaas van Aarsen

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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

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Apr 13, 2013
3,704
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:
 

Klaas van Aarsen

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Mar 5, 2012
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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)