This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ESvariant of any CPD set is a subset of an MC ∃ CPD 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 CPD, 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 ∃ CPD. We proved that Γ * is consistent. Let’s now prove that it’s maximally consistent. An ESvariant of Γ ∪ { ∼ P } is CPD ↓ Any ESvariant of Γ ∪ { ∼ P } ⊆ a MC ∃ CPD 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 CPD. Since P k is a PL sentence, it occurs kth in our enumeration. By the def. of our Γsequence, Γ k+1 = Γ k ⋃ { P k} if that’s CPD. Γ k ⋃ { P k} is CPD. 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 CPD. I.e. Γ * is maximal. An ESvariant of Γ ∪ { ∼ P } is CPD ↓ Any ESvariant of Γ ∪ { ∼ P } ⊆ a MC ∃ CPD 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 *....
View
Full
Document
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.
 Fall '10
 DerekAllen
 Philosophy

Click to edit the document details