I'll use the notion of 'satisfiable' (there is an interpretation in which all the members the set are true, and 'unsatisfiable' denoting the negation of that) rather than 'consistent' (there is no deduction of a contradiction from the members of the set, and 'inconsistent' denoting the negation of that), to keep the matter all semantical, and as it is an obvious and easy to show theorem that if a set of sentences is inconsistent then it is not satisfiable.
Here I changed some variables from previously, to avoid using T as both a variable and relation symbol, and to make the role of others more clear. Hope I don't make any typos:
We already have (1) below:
(1) Definition of 'is a valid argument':
For all g(g is a valid argument
if and only if
(g is an argument
and
for all i(if i is an interpretation, then it is not the case that
((for all p(if p is a premise of g, then p is true per i))
and
for all c(if c is the conclusion of g then c is false per i)))))
Symbolized:
Let Vx stand for x is a valid argument
Let Bx stand for x is an argument
Let Dx stand for x is an interpretation
Let Rxy stand for x is a premise of y
Let Txy stand for x is true per y
Let Uxy stand for x is the conclusion of y
Let Fxy stand for x is false per y
Ag(Vg
<->
(Bg
&
Ai(Di ->
~((Ap(Rpg -> Tpi))
&
Ac(Ucg -> Fci)))))
(2) Then we want to show that, for any argument g, if there is no interpretation in which all the premises of g are true, then g is valid:
Ag((g is an argument
&
~Ei(Di & Ap(Rpg -> Tpi))) -> Vg)
It's merely a tedious, routine exercise to do the proof in a system of the first order predicate calculus.
(3) Also, we want to show that, for any argument g, if there is no interpretation in which the conclusion is false, then g is valid:
Ag((g is an argument
&
~Ei(Di & Ac(Ucg -> Fpi))) -> Vg)
It's merely a tedious, routine exercise to do the proof in a system of the first order predicate calculus.
(4) And you want to also show that, for any argument g, if there is no interpretation in which all the premises of g are true, and there is no interpretation in which the conclusion of g is false, then g is valid.
But that is implied, a fortiori, from (2) and (3), given this theorem of sentential logic:
((P -> Q) & (H -> Q)) -> ((P & H) -> Q)