Dark Mode

370 views

2 votes

Here are some very useful ways of characterizing propositional formulas. Start by constructing a truth table for the formula and look at the column of values obtained. We say that the formula is:

- satisfiable if there is at least one $T$
- unsatisfiable if it is not satisfiable, i.e., all entries are $F$
- falsifiable if there is at least one $F$
- valid if it is not falsifiable, i.e., all entries are $T$

If $F1, F2$ and $F3$ are propositional formulae/expressions, over the same set of propositional variables, such that $(F1\rightarrow F2)$ is falsifiable, $(F1\rightarrow F3)$ is Tautology, and $(F3\rightarrow F1)$ is invalid.

then which of the following is/are Necessarily false:

- $F2$ is tautology
- $F3$ is tautology
- $F1$ is a contingency.
- $F1\wedge F3$ is contradiction.

0

@Sampanna Nag to be falsifiable there must be atleast one value $False$. For $F1\rightarrow\ F2$ to be falsifiable, at least one of the row has a value of $(True, False)$. But to satisfy $(F3\rightarrow F1)$ as invalid, atleast one value of F1 must be $False$. Hence $F1$ is a contingency, which makes option C correct.

P.s. See my answer. It will be clear.

2

5 votes

Answer : **A,D**

Given :

- f1 → f2 is falsifiable ie for
**at least one row**value**(f1,f2)**is**(T,F)**. - f1 → f3 is tautology ie
**no row**value**(f1,f3)**is**(T,F)**. - f3 → f1 is invalid ie for
**at least one row**value**(f3,f1)**is**(T,F)**.

option A : from above 1^{st} point we know at least one row of f2 is False. Therefore, f2 is not tautology.

option B : if f3 is tautology then also all given statements hold.

option C : if f1 is contingency then also all given statements hold.

option D : from above 1^{st} point we know at least one row of f1 is True and from above 2^{nd} point we know when f1 is True, f3 must be True. Thus, we have value for at least one row (f1,f3) as (T,T). Thus, f1 ^ f3 is not contradiction.

2 votes

**Ans: Option A,D**

We know that the value of $(p\rightarrow q)$ is $False$ only for the case when $p$ is $True$ and $q$ is $False$.

- $(F1\rightarrow F2)$ is Falsifiable. It means that in the truth table at least one of the row has a value of $(True, False)$. Hence the
**option A which says F2 is a Tautology is definitely False.**

- $(F1\rightarrow F3)$ is Tautology, Hence no value is $(True, False)$.

An argument that is not valid is said to be "invalid". Also, Invalid implies that the argument is not true for all instances.

- It’s given, $(F3\rightarrow F1)$ is Invalid. Hence atleast one row has a value of $(True, False)$.

From 1,2,3 we can be sure that $F1$ is a contingency, as it can take both values of $True$ or $False$. Hence **Option C is True. **

For **Option B, **there is no problem even if $F3$ is a Tautology. Hence **Option B is not necessarily False. **

For **Option D, **$F1\wedge F3$ **cannot be contradiction** because there is atleast one $True$ value for both $F1$ and $F3$ (from 1 and 3). Hence **Option D is also necessarily False.**

discrete mathematics - Invalidity of propositional formula - Mathematics Stack Exchange

1.3.1 Valid and Invalid Argument Forms (ornl.gov)