# Modus Ponens on A Deduction?

1. Nov 11, 2013

### darkchild

P is a one-place predicate, t is a constant, v is a variable of P. P(t/v) denotes replacing v by t in P.

In the proof of a theorem, it is given that

Δ$\vdash$$\forall$vP.
(meaning $\forall$vP is deduced from the set of statements Δ.)

There exists an axiom scheme

$\vdash$\forall[/itex]vP$\rightarrow$(P(t/v).

Then modus ponens is applied to these two to prove that

Δ$\vdash$P(t/v).

I've never seen modus ponens applied to a deduction and it is used with so I scarcely know what to ask...how is this permissible? How does it work...same as regular modus ponens? Is there a proof that this is shows this is the same as modus ponens, or a definition that describes it?

2. Nov 12, 2013

### darkchild

The axiom scheme should be

$\vdash\forall$v→P(t/v).

Also, the last paragraph of my original post should say "it is used without explanation, so I scarcely know what to ask." Meaning give me whatever relevant information you've got.

3. Jan 5, 2014

### MLP

This is a day late and a dollar short since your question is from November but ...

I think the axiom scheme should be ∀vP→P(t/v) so we are given the following as premises:

∀vP
∀vP→P(t/v)

notice that this has the form
1. A
2. A→B

Modus ponens is the rule that says that given 1 and 2 we may infer B, i.e., P(t/v)