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

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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 *. We set out a procedure for constructing *, and proved that * was consistent. But theres 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 * included this clause: IF i { P i} is C-PD, and P i is oF the Form ( x ) Q : i+1 is i { P i, Q(a/x) }, where a is the alphabetically earliest constant not occurring in P i or any member oF i I asked, how do we know we can add Q(a/x) to k { P i}? The answer someone gave, and I agreed with, relied on the thought that fnitely constants appear in any i and P i. We were wrong. Last time Our main goal is to show that if P then P . We do *not* assume initially that is Fnite --- thats why compactness is an interesting result. So we do *not* assume that { P } is Fnite either! So might contain inFnitely many constants. Last time Now, in our construction of *, we sometimes need to add Q(a/x) to k, where a is a constant that isnt in any sentences in the set. How do we know this is possible? The answer explains why we bother transforming { P } into an evenly subscripted set. Lets now turn back to proving that our construction of * yields a set thats MC- C-PD. We proved that * is consistent. Lets now prove that its maximally consistent. An ES-variant of { P } is C-PD Any ES-variant of { P } a MC- C-PD set * (11.4.4) Goal: Show that * is maximally consistent. Suppose the contrary: Theres a P k * s.t. * { P k} is C-PD. Since P k is a PL sentence, it occurs kth in our enumeration. By the def. of our -sequence, k+1 = k { P k} if thats C-PD. k { P k} is C-PD. Since if { P k} were inconsistent with k, it would be inconsistent with every superset of k, e.g. *. So k+1 = k { P k} (...perhaps plus a substitution instance) But that means P k k+1, so because k+1 *, P k *. Contradiction. So theres no P k * s.t. * { P k} is C-PD. I.e. * is maximal. An ES-variant of { P } is C-PD Any ES-variant of { P } a MC- C-PD set * (11.4.4) Finally, lets show that * is existentially complete. I.e., that for each sentence in * of the form ( x)Q , theres a substitution instance of ( x)Q in *. So suppose ( x)Q is in *....
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MIT24_241F09_lec24 - Logic I - Session 24 Completeness of...

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