Another Round With the Opponent on the Necessity of Identity

The Opponent writes,

The Maverick Philosopher has a comment on my earlier question about the necessity of identity. Can we get from ‘a=b’ to ‘necessarily a=b’ in a simple step? He thinks we can.

Now if ‘H’ and ‘P’ designate one and the same entity, then what appears to be of the form a = b, reduces to the form a = a. Clearly, if a = a, then necessarily, a = a. The assumption that the identity of H with P is contingent entails the absurdity that a thing is distinct from itself. Therefore the relation denoted by ‘=’ holds necessarily in every case in which it holds. Q. E. D.

The problem is the claim that ‘H’ (‘Hesperus’) and ‘P’ (‘Phosphorus’) designate one and the same entity. How do we get there, given only that H is the same object as P? Suppose we grant that H and P are this ‘one and the same entity’. We are saying that there is some entity, call it ‘V’ (i.e. Venus), such that H is identical with V and P is identical with V. Fair enough. But how do we get from there to the claim that the names designate this one and the same entity, i.e. that ‘H’ designates V and ‘P’ designates V? I.e. what validates the move from 2 to 3 in the following argument?

1. H=V
2. ‘H’ designates H
3. Therefore ‘H’ designates V.

You need the principle of substitutivity, the principle that if a=b and Fa, then infer Fb. For example, let F be the function ‘‘H’ designates –’. Then we agree that F(H), because we assumed that ‘H’ designates H. And we posit that H=V. Given substitutivity, it follow that F(V). But only given that substitutivity is valid in this case, which is not at all obvious, at least to me.

RESPONSE

I am afraid I just don't understand what the Opponent's problem is.  He writes, "The problem is the claim that ‘H’ (‘Hesperus’) and ‘P’ (‘Phosphorus’) designate one and the same entity. How do we get there, given only that H is the same object as P?"  Apparently, the Opponent wants to know what validates the inference from

Hesperus is the same entity as Phosphorus

to

'Hesperus' and 'Phosphorus' designate the same entity.

What validates the inference is the principle that if two putatively distinct entities are in fact numerically the same entity, then the names for these putatively distinct entities are co-referential: they designate one and the same entity.

I don't see the need to invoke a principle of substitutivity.  In the above inference there was no substitution of a name for a name.   

Comments

[Astute opponent]

Fortunately this is purely a question of logic, with minimal philosophical content.

>>if two putatively distinct entities are in fact numerically the same entity, then the names for these putatively distinct entities are co-referential: they designate one and the same entity.

Simplifying, where n(name, bearer) is the naming function:

Principle: (x)(y) [ x=y & n(‘x’,x) -> n(‘x’,y) ]

Thus
1. a=b
2. n(‘a’,a)
3. n(‘a’,b)

The move from 2 to 3 invokes your principle. Note the substitution of ‘b’ for ‘a’.

>>I don't see the need to invoke a principle of substitutivity. In the above inference there was no substitution of a name for a name.

But your inference involved the principle I formalised above, which is a specific instance of the principle of substitutivity. This does indeed require the substitution of a name for a name.

To win this game you need to formalise what you said above in a way that does not invoke substitutivity.