2
$\begingroup$

Use a truth table to show that the following argument form is invalid.

p$\rightarrow$q

q

∴p

My attempt: I made a truth table as below, but in what column and row can I find the argument form is invalid?

$$\begin{array}{c|c|c|c} \space\space\space\space(p \rightarrow q)&\land &q& \rightarrow& p \\\hline T\space T\space T&\space T&\space T&T&T \\TFF&F&F&T&T\\F\color {red} {T}T&T&\color {red} {T}&F&\color {red} {F}\\FTF&F&F&T&F \\\hline step 1\space2\space 1&\space3&\space1&\space4&\space1 \end{array}$$

[EDIT] I think it's found invalid in the row where I wrote in red since it shows the conclusion p is false when all the premises (p$\rightarrow$q) and $q$ are true.

FYI ""A row of the truth table in which all the premises are true is called a critical row. ...You can show that an argument is invalid by constructing a truth table for the argument form and finding at least one critical row in which all the premises are true but the conclusion is false."

Source: Discrete Mathematics with Applications by Susanna Epp

$\endgroup$
  • 2
    $\begingroup$ Sorry, was confused by your truth table. That's not how we wrote 'em when I was a young man. Your truth table shows exactly when it is false - when $p$ is false and $q$ is true, then $p\implies q$ is true, $q$ is true, but $p$ is false. $\endgroup$ – Thomas Andrews Jan 14 '16 at 14:09
  • $\begingroup$ math.stackexchange.com/help/someone-answers $\endgroup$ – 1 0 Jan 14 '16 at 14:43
2
$\begingroup$

Note that the question is not about the formula $$(p\to q)\land q \rightarrow p $$ but about the proposed inference $$ (p\to q), q \vdash p $$ This is important because $\vdash$ is not a connective and shouldn't have a truth value.

In order to use a truth table, as requested by the question, you should have "input" columns $p$ and $q$, and "output" columns $p\to q$ as well as $q$ and $p$:

$$ \begin{array}{cc|ccc} p & q & p\to q & q & p \\ \hline F & F & T & F & F \\ T & F & F & F & T \\ F & T & \cdots \\ T & T & & \cdots \end{array} $$

If you can find a row where the $p \to q$ and $q$ outputs are both T and the $p$ output is $F$, then your truth table shows that the inference $(p\to q), q\vdash p$ is not valid.

$\endgroup$
2
$\begingroup$

As not every row in $\to $ is T, then the argument is invalid ($(p \to q) \land q\to p $ is not a tautology).

$\endgroup$
  • $\begingroup$ What is "it" in "it is not a tautology"? $\endgroup$ – Thomas Andrews Jan 14 '16 at 14:07
  • $\begingroup$ @Thomas Andrews, solved. $\endgroup$ – JnxF Jan 14 '16 at 14:24
1
$\begingroup$

With fewer typing effort:

The implication $P \implies Q$ is true when $P$ is false. Now, if in addition $Q$ is true, then

$$\left(\left(P \implies Q\right) \land Q\right) \implies P$$

is false. Therefore, in general, the logical expression

$$\left(\left(P \implies Q\right) \land Q\right) \implies P$$

is not a tautology. Hence, the argument form is invalid.

$\endgroup$
1
$\begingroup$

$$((p\implies q)\land(q))\implies p$$ This is contingency but not valid. Counter example : $\text{put, p=False, and q = True }$

Then LHS is True but RHS is False, which means implication (i.e, $ \implies $) is False.

$$((F\implies T)\land T)\implies F$$

$$(T\land T)\implies F$$

$$T\implies F$$

$$F$$

Using truth table, we can see : This formula is not tautology, but it's contingency.

enter image description here

A formula of propositional logic is a tautology if the formula itself is always true regardless of which valuation is used for the propositional variables.

But, we have not all True value on the above table. So, it violates the definition of tautology.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.