@Reconstructo,

Reconstructo;154712 wrote:We have the 1 and the 0 and the NOT. Let's exclude the other operators. Is this the minimal dynamic math/logic system?

I'm not clear as to what you mean here but my impression is that you want to produce 'propositional logic' with the minimum of primitive terms and to use a minimal logical arithmetic to demonstate the theorems of this system.

We cannot produce the system of propositional logic with [1,0, ~], but we can produce the system with [1,0,|].

Let (p, q, r, ...) be propositional variables which take on the values 1 and 0 (truth and falsity).

(nor) primitive (|): 1|1=0, 1|0=0, 0|1=0, 0|0=1.

(not)

~p = p|p.

~1=1|1, ~0=0|0.

ie: ~1=0 and ~0=1.

(or)

(p v q) = ~(p|q).

(1 v 1)=~(1|1), (1 v 0)=~(1|0), (0 v 1)=~(0|1), (0 v 0)=~(0|0).

ie: (1 v 1)=1, (1 v 0)=1, (0 v 1)=1, (0 v 0)=0.

(if then)

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

(1 -> 1)=(~1 v 1), (1 -> 0)=(~1 v 0), (0 -> 1)=(~0 v 1), (0 -> 0)=(~0 v 0).

ie: (1 -> 1)=1, (1 -> 0)=0, (0 -> 1)=1, (0 -> 0)=1.

(and)

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

(1 & 1)=~(~1 v ~1), (1 & 0)=~(~1 v ~0), (0 & 1)=~(~0 v ~1), (0 & 0)=~(~0 v ~0).

ie: (1 & 1)=1, (1 & 0)=0, (0 & 1)=0, (0 & 0)=0.

(equivalence)

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

(1 <-> 1)=((1 -> 1) & (1 -> 1)), (1 <-> 0)=((1 -> 0) & (0 -> 1)),

(0 <-> 1)=((0 -> 1) & (1 -> 0)), (0 <-> 0)=((0 -> 0) & (0 -> 0)).

ie: (1 <-> 1)=1, (1 <-> 0)=0, (0 <-> 1)=0, (0 <-> 0)=1.

With this simple arithmetic we can show the theorems (tautologies) of the system.

For example: (p & (p-> q)) -> q.

From the above we get:

(1 & (1 -> 1)) -> 1, (1 & 1) -> 1, 1 -> 1, 1.

(0 & (0 -> 1)) -> 1, (0 & 1) -> 1, 0 -> 1, 1.

(1 & (1 -> 0)) -> 0, (1 & 0) -> 0, 0 -> 0, 1.

(0 & (0 -> 0)) -> 0, (0 & 1) -> 0, 0 -> 0, 1.

That is, (p & (p -> q)) resolves to 1 for all values of p and q, therfore,

(p & (p -> q)) -> q, is a theorem.