Which of the following expressions are formulae of statement logic? My task is: which of the following expressions are formulae of statement
logic? Justify your answer. If an expression is a formula, in which brackets
are omitted, then rewrite this formula with all the brackets.
1. V (p ∨ ¬p)
2. (0 ∧ 1) →  (1 ∨ 0)
3. ƛx.x + x ≠ x
4. (¬(p → q ∧ ¬q → p))
5. p ∨ q ∨ r → p ∧ q ∨ r
6. (((p → q) ∧ p → q)
7. ((p → p) → p) → p → p
8. ¬(p → ¬(q ↔ ¬p))

According to the rules I have found:
Syntactic Rules: 


*

*Any atomic statement is a formula. 

*If ϕand ψare formulas then ¬ϕ, (ϕ ∧ ψ), (ϕ ∨ ψ), (ϕ → ψ) and (ϕ ↔ ψ) are also 
formulas. 

*There are no other formulas (in Statement Logic). 


i can guess:
1. not a formula
2. not a formula
3. not a formula
4. (¬(p → q) ∧ (¬(q → p)))
5. (p ∨ q) ∨ r → p ∧ (q ∨ r)
6. (p → q) ∧ (p → q) 
7. ((p → p) → p) → p → p
8. ¬(p → ¬(q ↔ ¬p))

 A: You’ve got some of it, but there are some mistakes. $(1),(2),(3)$, and $(8)$ are correct.
$(4)$ is probably intended to be $\neg((p\to q)\land(\neg q\to p))$, not $(\neg(p\to q)\land\neg(q\to p))$, which is the corrected version of your answer. You have an unwanted pair of parentheses around $\neg(q\to p)$: note that $\neg\varphi$ doesn’t require surrounding parentheses. If it’s really ill-formed, it might be intended to be $(\neg(p\to q)\land(\neg q\to p))$.
$(5)$ requires more parentheses: every binary connective together with its arguments is surrounded by a pair. There are different ways to associate the expressions; the one closest to what you have is $\big(((p\lor q)\lor r)\to(p\land(q\lor r))\big)$.
$(6)$ is almost right: you’re just missing the parentheses surrounding the whole thing, to match the $\land$ in the middle. Make it $((p\to q)\land(p\to q))$.
In $(7)$ you’re missing two required pairs of parentheses: each of the implications gets a pair, so you should have $((((p\to p)\to p)\to p)\to p)$, though $(((p\to p)\to p)\to(p\to p))$ is another possible interpretation.
