428 views

If $F1$, and $F2$ are propositional formulae/expressions, over same set of propositional variables, such that $F1,F2$ both are contingencies, then which of the following is/are necessarily false(i.e. Never Possible):

1. $F1 \vee F2$ is a contingency.
2. $F1 \vee F2$ is a tautology.
3. $F1 \vee F2$ is a contradiction
4. $(F1 \rightarrow F2) \vee (F2 \rightarrow F1)$ is contingency.

@Deepak Poonia Thank you Sir. I understood my mistake.

Since F1 and F2 are both contingencies, so for some situation F1 and F2 both can be false..then F1vF2 can be a contradiction..so why is option C not true ?

@Sampanna Nag

Since $F_1$ is a Contingency, so, for some row $r$ in the truth table of $F_1.$ $F_1$ is True.

So, for this row $r,$ $F_1 \vee F_2$ is also True. Hence, $F_1 \vee F_2$ is True for at least one row, hence, can Never be a Contradiction.

If $F1$ and $F2$ are contingencies.

1. This is obvious that it is not necessarily false since both $F1$ and $F2$ are contingency, $F1 \lor F2$ is a contingency.
1. $F1 \lor F2$ is tautology is not necessarily false because conside the situation where either of $F1$ or $F2$ is true when other is false. This will give $F1 \lor F2$ a tautology.
2. $F1 \lor F2$ is a contradiction is necessarily false because to be a contraction $F1 \lor F2$ must always be false for that both $F1$ and $F2$ must be contradiction.
3. $(F1→F2)∨(F2→F1)$ = $\lnot F1 \lor F2 \lor \lnot F2 \lor F1$ is always tautology regardless what logical value $F1$ and $F2$ takes, so option (D) is also necessarily false.

by

### 1 comment

Correction in statement A:

1. This is obvious that it is not necessarily false since both F1 and F2 are contingency, F1∨F2 is a contingency.

Since $F_1, F_2$ both are contingency, So, $F_1 \vee F_2$ may be contingency, but it is Not necessary that $F_1 \vee F_2$ will definitely be contingency. For example, Let $F_1 = P, F_2 = \neg P,$ here, $F_1 \vee F_2$ is Tautology, not contingency.

Correction in statement B:

1. F1∨F2 is tautology is not necessarily false because conside the situation where either of F1 or F2 is true when other is false. This will give F1∨F2 a tautology.

Since $F_1, F_2$ both are contingency, So, $F_1 \vee F_2$ may be Tautology, but it is Not necessary that $F_1 \vee F_2$ will definitely be a Tautology. For example, Let $F_1 = P, F_2 = \neg P,$ here, $F_1 \vee F_2$ is Tautology, but if  $F_1 = P, F_2 = P,$ here, $F_1 \vee F_2$ is Not Tautology.

Another example, $F_1 = P\vee Q, F_2 = P \rightarrow Q,$ here, $F_1 \vee F_2$ is Tautology