Could you please point out the flaw in my argument?

Dark Mode

6,010 views

37 votes

Best answer

$$\begin{array}{|c|c|c|c|c|} \hline \textbf{p} & \textbf{q}& \textbf{p} \leftrightarrow \textbf{q} &\textbf{p} \to \lnot \textbf{q} & \begin{align*}(\boldsymbol{p} \leftrightarrow \boldsymbol{q}) & \to (\boldsymbol{p}\to \neg \boldsymbol{q}) \end{align*}\\\hline \text{0} & \text{0}& \text{1} & \text{1} & \text{1}\\\hline \text{0} & \text{1}& \text{0} & \text{1} & \text{1}\\\hline \text{1} & \text{0}& \text{0} & \text{1} & \text{1}\\\hline \text{1} & \text{1}& \text{1} & \text{0} & \text{0} \\\hline \end{array}$$

So, "imply" is FALSE making does not imply TRUE.

So, "imply" is FALSE making does not imply TRUE.

0

1

0

0

but truth table obeys the truth table of implication truth table.

p | q | $p\rightarrow q$ |

T | T | T |

T | F | F |

F | T | T |

F | F | T |

from this table if we say that "p implies q" even when we can give a combination in which $p\rightarrow q$ is false.

like this, $(p\leftrightarrow q) \rightarrow (p\rightarrow \sim q)$ is false only when $(p\leftrightarrow q)$ is T & $(p\rightarrow \sim q)$ is F.

rest of the cases it will true. It looks like an implication truth table.

0

0

I think both the followings are different -

- prove that $(p\Leftrightarrow q)\rightarrow (p\rightarrow \sim q)$ is a tautology.
- prove that $(p\Leftrightarrow q)\rightarrow (p\rightarrow \sim q)$.

In 1st one if you show a combination of p,q such that LHS is T & RHS is F, you can declare that this is not a tautology.

In 2nd one you have to check whether LHS implies RHS or not, i.e you've to check whether this follows the implication rule(means truth table) or not.

Now according to the qsn. language, it feels like they want the 2nd one, not the 1st one.

0

@MRINMOY_HALDER can you tell me whether this statement is correct or wrong

"if I say this p⟺q implies p→¬q and i say it as true then for every combination of p and q it should be true"

0

0

@MRINMOY_HALDER so basically this question is asking whether (*p*⇔*q*)→(*p*→∼*q*) is a tautology or not, isn't it?

0

edited
Mar 25, 2021
by subbus

In last column of above truth table $p \Longleftrightarrow q \rightarrow p \to \lnot q$ doesn’t have all 1’s.Therefore, it is not a tautology

If It is a tautology(all 1’s in last column) then $p \Longleftrightarrow q$ implies $p \to \lnot q$

i.e., $ \left( p \Longleftrightarrow q~~~ \rightarrow ~~~ p \to \lnot q \right) \equiv True $

But Now it is not tautology so $p \Longleftrightarrow q$ does not implies $p \to \lnot q$

ie. $\left( p \Longleftrightarrow q ~~~ \not \rightarrow ~~~p \to \lnot q \right) \equiv True$

If It is a tautology(all 1’s in last column) then $p \Longleftrightarrow q$ implies $p \to \lnot q$

i.e., $ \left( p \Longleftrightarrow q~~~ \rightarrow ~~~ p \to \lnot q \right) \equiv True $

But Now it is not tautology so $p \Longleftrightarrow q$ does not implies $p \to \lnot q$

ie. $\left( p \Longleftrightarrow q ~~~ \not \rightarrow ~~~p \to \lnot q \right) \equiv True$

1