Question

Its for soundness theorem. I need to prove that the axioms (∀x)A -> A[t/x] its valid in constant domain semantics.

I assume there's a world in an arbitrary model within (∀x)A -> A[t/x] it's false and i search for a contradiction.
(∀x)A -> A[t/x] its false when (∀x)A its true and A[t/x] its false. (∀x)A its true when A is true for each object o in U (or D, the domain) that the valuation function v assigns to the variable, but I don't understand how to explain that A[t/x] its false in logic syntax.

Answer 1

In constant domain semantics, the axiom (∀x)A -> A[t/x] is valid. It is true when (∀x)A is true and A[t/x] is true, and false when (∀x)A is true and A[t/x] is false.

Step-by-step explanation:

The soundness theorem is a fundamental theorem in mathematical logic that relates to the validity of logical formulas. In constant domain semantics, we can prove that the axiom (∀x)A -> A[t/x] is valid.

To prove this, let's assume that (∀x)A is true. This means that A is true for every object x in the domain U. Now, substituting t for x in A[t/x], we get A[t/x]. Since A is true for every object x, it must also be true for t. Therefore, (∀x)A -> A[t/x] is valid.

On the other hand, (∀x)A -> A[t/x] is false when (∀x)A is true and A[t/x] is false. This occurs when A is true for every object x in the domain U, but A[t/x] is false for some object t in U. In logic syntax, A[t/x] represents the formula A with the term t substituted for the variable x. If this substitution leads to a false statement, then (∀x)A -> A[t/x] is false.