If we grant that there are 4 truth values: T = logical truth, t = empirical truth, f =empirical falsity, F = logical falsity.

Not:

~T=F, ~t=f, ~f=t, ~F=T.

Or:

TvT=T, Tvt=T, Tvf=T, TvF=T, tvT=T, tvt=t, tvf=T, tvF=t

fvT=T, fvt=T, fvf=f, fvF=f, FvT=T, Fvt=t, Fvf=f, FvF=F.

Necessity:

[]T=T, []t=F, []f=F, []F=F.

Definitions:

<>p =df ~[]~p.

(p -> q) =df ~p v q.

(p & q) =df ~(~p v ~q).

(p <-> q) =df ((p -> q) & (q -> p)).

All of the theorems of the 2-valued logic are tautologous here.

The aditional axioms of S5 are given..

http://mally.stanford.edu/S5.html
1. [](p -> q) -> ([]p -> []q).

2. []p -> p.

3. <>p -> []<>p.

2. []T -> T, []t -> t, []f -> f, []F -> F. Each are tautologous by the above.

3. <>T -> []<>T, <>t -> []<>t, ,<>f -> []<>f, <>F -> []<>F. Each of these are also tautologous.

1. p q | [](p -> q) -> ([]p -> []q))

....T.T...T.....T.......T....T....T..T

....t..T...T.....T.......T....F....T..T

....f..T...T.....T.......T....F....T..T

....F.T...T.....T.......T....F....T..T

....T.t....F.....t.......T....T....F...F

....t..t...T.....T.......T....F....T...F

....f..t...F......t.......T....F....T...F

....F..t...T.....T.......T....F....T...F

....T..f...F.....f........T....T....F...F

....t...f...F.....f........T....F....T...F

....f...f...T.....T.......T....F....T...F

....F..f...T.....T.......T....F....T...F

....T..F...F.....F.......T....T....F...F

....t..F....F.....f.......T....F....T...F

....f..F....F.....t.......T....F....T...F

....F..F...T.....T......T....F....T...F

That is, all of the axioms of S5 are tautologies.

Doesn't it follow that all theorems (tautologies) of S5 are show to be the case by this method?