In my textbook of the propositional logic I should indentify all subformulas of the following formula: ¬(a∧b)⟺(c⇒a).
According to the textbook the formula (a∧b)⟺(c⇒a) is not a subformula of the formula ¬(a∧b)⟺(c⇒a). Why?
They also argue that ¬(a∧b)⟺(c⇒a) is a subformula of itself, is that correct?
On
If the statement was ¬[(a∧b)⟺(c⇒a)] (i.e. if the statement was the negation of a biconditional), then (a∧b)⟺(c⇒a) would indeed be a subformula. But, the given formula ¬(a∧b)⟺(c⇒a) is a biconditional between ¬(a∧b) and c⇒a. So, those are both subformulas, but (a∧b)⟺(c⇒a) is not
Keep in mind that a subformula is not the same as a substring! Rather, try and understand how the operations are applied to the parts: that will tell you waht the subformulas are.
And yes, any formula is a subformula of itself ... kind of like how every set is a subset of itself. If we want to exclude the formula itself, we would call for strict subformulas.
The formula $\lnot(a \land b) \iff (c \Rightarrow a)$ can be ambiguous if we do not assume a convention about omitting parentheses.
According to the usual convention, the negation symbol applies to as little as possible.
This means that the above formula is an abbreviation for :
Regarding the second part of the question, the answer is YES: a formula is a sub-formula of itself, as must be evident reviewing the definition of sub-formula in your textbook.
Subformulas can be "mechanically" generated from the original formula (which is a sub-formula of itself) removing the main connective (there is only one).
Thus, from $(\lnot(a \land b)) \iff (c \Rightarrow a)$ we have to remove $\iff$ to produce two new sub-formulas :
Now we apply the same procedure to $\lnot(a \land b)$ removing the negation sign to produce the new sub-formula $a \land b$ and again to get : $a$ and $b$.
The same for $(c \Rightarrow a)$.