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Modus Ponens on A Deduction?

  1. Nov 11, 2013 #1
    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[itex]\vdash[/itex][itex]\forall[/itex]vP.
    (meaning [itex]\forall[/itex]vP is deduced from the set of statements Δ.)

    There exists an axiom scheme

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

    Then modus ponens is applied to these two to prove that

    Δ[itex]\vdash[/itex]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. jcsd
  3. Nov 12, 2013 #2
    The axiom scheme should be

    [itex]\vdash\forall[/itex]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.
     
  4. Jan 5, 2014 #3

    MLP

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    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)
     
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