If formula $\phi$ proves a contradiction $\bot$ then do we have $\vdash\phi\to\bot$? I am trying to teach myself some logic by means of "A Friendly Introduction to Mathematical Logic" of Leary and Kristiansen.
It has a focus on formulas in the sense that axioms are not necessarily sentences.
Interference rules practicized in that book are PC (propositional consequence) and the quantifier rule QR stating that from $\psi\to\phi$ we can deduce $\psi\to\forall x\phi$ if $x$ is not free in $\psi$.

Let $\mathcal{L}$ be a first order language, let $\bot$ denote some
$\mathcal{L}$-sentence of the form $\psi\wedge\neg\psi$ and let
$\phi$ be an $\mathcal{L}$-formula.
Then $\Sigma:=\left\{ \phi\right\} $ is by definition inconsistent
if there is a deduction from $\Sigma$ to $\bot$.
Now my question:

If $\left\{ \phi\right\} $ is inconsistent then can it be proved that also: $\vdash\phi\to\bot$?

It is clear to me that the answer is "yes" if $\phi$ is a sentence because then we can apply the deduction theorem.
But what if $\phi$ is not a sentence?

My try:
If $\tilde{\phi}$ denotes a universal closure of $\phi$ then $\left\{ \tilde{\phi}\right\} \vdash\phi$
so that by transitivity of $\vdash$ we have $\left\{ \tilde{\phi}\right\} \vdash\bot$
and appying deduction theorem we have $\vdash\tilde{\phi}\to\bot$.
But this only shifts the problem to another question:

If there is a deduction $\vdash\tilde{\phi}\to\bot$ then is there also a deduction $\vdash\phi\to\bot$?


Thank you in advance and my apologies if this question is a duplicate.
 A: I didn't read Leary and Kristiansen book but I'm currently reading "Introduction to mathematical logic" from Mendelson so i hope i can answer your first question.
As for your first question that states "if a formula 
ϕ
 (being 
ϕ
:
ψ
∧
¬
ψ
)  proves a contradiction 
⊥
 then do we have 
⊢
ϕ
→
⊥
?" I can answer: effectively a formula that states 
ψ
∧
¬
ψ
 is going to make us conclude 
⊥
, this ( ψ ∧ ¬ ψ ) → ⊥ formula is a theorem, here's a proof of the formula using natural deduction rules:
1)
ψ
∧
¬
ψ
 - assumtion
2)
ψ
 - rule E∧ in 1
3)¬
ψ
 - rule E∧ in 1
4)
⊥
 - in 2,3
5)(
ψ
∧
¬
ψ
) → 
⊥
 - rule I→ in 1,4
If we look at the truth table of 
ψ
∧
¬
ψ
 all values are false, this means that, not only ( ψ ∧ ¬ ψ ) → ⊥ is a tautology, but also (
ψ
∧
¬
ψ
) → X (where X is an arbitrary formula) is a tautology. If a formula proves a contradiction u have probe all the formulas.
As for your second question I'm not confident enought to give an answer, im currently reading chapter two of Mendelson's book (first-order logic) so I share your doubt.
