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Consider two well-formed formulas in propositional logic

$F_1: P \Rightarrow \neg P$          $F_2: (P \Rightarrow \neg P) \lor ( \neg P \Rightarrow P)$

Which one of the following statements is correct?

1. $F_1$ is satisfiable, $F_2$ is valid
2. $F_1$ unsatisfiable, $F_2$ is satisfiable
3. $F_1$ is unsatisfiable, $F_2$ is valid
4. $F_1$ and $F_2$ are both satisfiable

$F_1: P \Rightarrow \neg P\equiv\neg P\vee \neg P\equiv\neg P$

When we put $P\equiv T$ it will $F$

and When we put $P\equiv F$ it will $T$

It is called contingency.

Always false called contradiction or unsatisfiable

Always true called valid or tautology

At least one true called satisfiable.

$F_2: (P \Rightarrow \neg P) \lor ( \neg P \Rightarrow P)\equiv\neg P\vee P\equiv T$(Always true)

Correct ME

F1: P => ~p

This statement have logical implication(=>) means  P → ~p is tautology

therefore F1 is always true (therefore F1 is valid statement)
If this question would be MSQ, then option A) and D), both are correct.

$F1: P\to \neg P$

$=\neg P\vee \neg P$

$=\neg P.$      can be true when P is false ( Atleast one T hence satisfiable)

$F2: (P\to \neg P)\vee (\neg P\to P)$

$=\neg P \vee (P\vee P)$

$=\neg P \vee P$

$=T.$

VALID

Option (A)

Read the entire comment and answer, you will understand better.

ok

Arjun Sir, there's some typo here in F1

=¬P¬P

it should be -P v –P

"A valid (true for every set of values)  formula is always satisfiable (true for at least one value)  , but a satisfiable formula may or may not be valid".

as F1 is not valid but satisfiable,

and F2 is valid(which implicitly means is satisfiable too).

therefore OPTION A is the complete solution, but OPTION D is not.

by
I can see both are satisfiable i wl go for option b

valid means always true. That is, whatever be the values assigned to the variables, the formula returns true. eg. $p\vee \neg p$
(a) is the most appropriate.

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