This is from a text on mathematical logic. The theorem to be proven (specialization):(adsbygoogle = window.adsbygoogle || []).push({});

If Δ [itex]\vdash[/itex] [itex]\forall[/itex]vP, then Δ [itex]\vdash[/itex]P(t/v), provided thatPadmitstforv.

My confusion concerns the use of modens ponens in the proof:

Suppose that Δ [itex]\vdash[/itex] [itex]\forall[/itex]vPandPadmitstforv. Then modus ponens applied to Δ [itex]\vdash[/itex] [itex]\forall[/itex]vPand [itex]\vdash \forall[/itex]vP[itex]\rightarrow[/itex]P(t/v)(Axiom Scheme A5) gives Δ [itex]\vdash[/itex]P(t/v).

I have never seen this before and do not understand how it is legal or exactly what it means to use modus ponens on statements containing Δ (a set of formulas used as premises) and [itex]vdash[/itex]. It seems the latter are simply ignored, yet they are crucial to the meaning of the statement.

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# Modus Ponens on A Statement of Deduction?

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