MIT24_241F09_lec24

# MIT24_241F09_lec24 - Logic I Session 24 Completeness of PD...

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Unformatted text preview: Logic I - Session 24 Completeness of PD Last time We started to prove that PD is complete. We were working on proving that an ES-variant of any C-PD set is a subset of an MC- ∃ C-PD set Γ *. We set out a procedure for constructing Γ *, and proved that Γ * was consistent. But there’s one thing I need to correct... Last time Alex asked last time why we need to deal with evenly subscripted sets. In my answer, I misidentifed the reason. Our procedure For building Γ * included this clause: IF Γ i ∪ { P i} is C-PD, and P i is oF the Form ( ∃ x ) Q : Γ i+1 is Γ i ∪ { P i, Q(a/x) }, where a is the alphabetically earliest constant not occurring in P i or any member oF Γ i I asked, how do we know we can add Q(a/x) to Γ k ∪ { P i}? The answer someone gave, and I agreed with, relied on the thought that fnitely constants appear in any Γ i and P i. We were wrong. Last time Our main goal is to show that if Γ ⊨ P then Γ ⊢ P . We do *not* assume initially that Γ is Fnite --- that’s why compactness is an interesting result. So we do *not* assume that Γ ⋃ { ∼ P } is Fnite either! So Γ might contain inFnitely many constants. Last time Now, in our construction of Γ *, we sometimes need to add Q(a/x) to Γ k, where a is a constant that isn’t in any sentences in the set. How do we know this is possible? The answer explains why we bother transforming Γ ⋃ { ∼ P } into an evenly subscripted set. Let’s now turn back to proving that our construction of Γ * yields a set that’s MC- ∃ C-PD. We proved that Γ * is consistent. Let’s now prove that it’s maximally consistent. An ES-variant of Γ ∪ { ∼ P } is C-PD ↓ Any ES-variant of Γ ∪ { ∼ P } ⊆ a MC- ∃ C-PD set Γ * (11.4.4) Goal: Show that Γ * is maximally consistent. Suppose the contrary: There’s a P k ∉ Γ * s.t. Γ * ⋃ { P k} is C-PD. Since P k is a PL sentence, it occurs kth in our enumeration. By the def. of our Γ-sequence, Γ k+1 = Γ k ⋃ { P k} if that’s C-PD. Γ k ⋃ { P k} is C-PD. Since if { P k} were inconsistent with Γ k, it would be inconsistent with every superset of Γ k, e.g. Γ *. So Γ k+1 = Γ k ⋃ { P k} (...perhaps plus a substitution instance) But that means P k ∈ Γ k+1, so because Γ k+1 ⊆ Γ *, P k ∈ Γ *. Contradiction. So there’s no P k ∉ Γ * s.t. Γ * ⋃ { P k} is C-PD. I.e. Γ * is maximal. An ES-variant of Γ ∪ { ∼ P } is C-PD ↓ Any ES-variant of Γ ∪ { ∼ P } ⊆ a MC- ∃ C-PD set Γ * (11.4.4) Finally, let’s show that Γ * is existentially complete. I.e., that for each sentence in Γ * of the form ( ∃ x)Q , there’s a substitution instance of ( ∃ x)Q in Γ *. So suppose ( ∃ x)Q is in Γ *....
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## This note was uploaded on 01/25/2012 for the course PHIL 201H1F taught by Professor Derekallen during the Fall '10 term at University of Toronto.

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MIT24_241F09_lec24 - Logic I Session 24 Completeness of PD...

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