in Mathematical Logic retagged by
2 votes
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:

  1. $F2$ is tautology
  2. $F3$ is tautology
  3. $F1$ is a contingency.
  4. $F1\wedge F3$ is contradiction.
in Mathematical Logic retagged by


If F1 is contingency then F1 can be either True or False. But if F1 is False then F1->F2 is never falsifiable. So how is option C correct?

@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.


@Abhrajyoti00 Thank you!


2 Answers

5 votes
5 votes

Answer : A,D

Given :

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

option A : from above 1st 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 1st point we know at least one row of f1 is True and from above 2nd 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
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 $False$. 

  1. $(F1\rightarrow F2)$ 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. 
  1. $(F1\rightarrow F3)$ 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.

  1. 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 (



Related questions