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.
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.