That’s it. Now override final would simply mean
This function overrides a base class one (override) and cannot be overridden itself (final).“
final on it’s own would impose a weaker requirement. override and final have independent behavior.

#include <iostream>
struct A
    virtual void foo();
    void bar();
    virtual ~A();
// member functions definitions of struct A:
void A::foo() { [std::cout]( << "A::foo();\n"; }
A::~A() { [std::cout]( << "A::~A();\n"; }
struct B : A
//  void foo() const override; // Error: B::foo does not override A::foo
                               // (signature mismatch)
    void foo() override; // OK: B::foo overrides A::foo
//  void bar() override; // Error: A::bar is not virtual
    ~B() override; // OK: `override` can also be applied to virtual
                   // special member functions, e.g. destructors
    void override(); // OK, member function name, not a reserved keyword
// member functions definitions of struct B:
void B::foo() { [std::cout]( << "B::foo();\n"; }
B::~B() { [std::cout]( << "B::~B();\n"; }
void B::override() { [std::cout]( << "B::override();\n"; }
int main() {
    B b;;
    b.override(); // OK, invokes the member function `override()`
    int override{42}; // OK, defines an integer variable
    [std::cout]( << "override: " << override << '\n';

You will need a virtual to use override