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Consider two well-formed formulas in propositional logic

$F1: P \Rightarrow \neg P$          $F2: (P \Rightarrow \neg P) \lor ( \neg P \Rightarrow P)$

Which one of the following statements is correct?

1. F1 is satisfiable, F2 is valid
2. F1 unsatisfiable, F2 is satisfiable
3. F1 is unsatisfiable, F2 is valid
4. F1 and F2 are both satisfiable
edited | 1.1k views
Every valid formula is satisfiable.

So A & D both are correct but A is more precise.

F1: $P\to \neg P$

$=\neg P\wedge \neg P$

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

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

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

$=\neg P \vee P$

$=T.$

VALID

Option A

edited
The concept behind this solution is:
a) Satisfiable
If there is an assignment of truth values which makes that expression true.
b) UnSatisfiable
If there is no such assignment which makes the expression true
c) Valid
If the expression is Tautology
Here, P => Q is nothing but –P v Q
F1: P => -P = -P v –P = -P
F1 will be true if P is false and F1 will be false when P is true so F1 is Satisfiable
F2: (P => -P) v (-P => P) which is equals to (-P v-P) v (-(-P) v P) = (-P) v (P) =
Tautology
So, F1 is Satisfiable and F2 is valid
Option (a) is correct.
Why not option D

both F1 and F2 are satisfiable
+1 vote
1."A formula is called Satisfiable if it is true at least one case "

2."A formula is called valid if it is true in all the cases."

3.Valid formula or tautology are the same things.

4.Satisfiable and Contradiction are two opposite argument.

5.If formula is Satisfiable can not be Contradiction and vice-versa
–1 vote
I can see both are satisfiable i wl go for option b
If both are satisfiable (d) must be the answer rt? Also, isn't F2 valid making (a) a better answer?
What is a valid argument? I dont know plz explain i know the satisfiable only
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.