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Math 486 Homework 3 Notes Paul Shafer February 19, 2008 Everyone did well this week. I only have a few minor points. Problem I.5.1 Most of you handled each case by writing, for example, “if V ( α β ) = T then V ( α ) = T or V ( β ) = T ”. You should also say why this answers the question by writing something like “and each case provides a path through the corresponding atomic tableau”. Problem I.6.1 The book should do a better job of explaining the order it is using when choosing the next en- try to reduce in the CST construction. When generating τ m +1 from τ m , we choose the length- lexicographically least entry which is not reduced on some noncontradictory path. Recall the lex- icographic ordering on strings of length n : σ < L τ if and only if σ ( i ) < τ ( i ) where i is the ﬁrst position at which σ and τ diﬀer. The length-lexicographic order < LL on 2 is deﬁned as follows: σ < LL τ ⇔ | σ | < | τ | ∨ ( | σ | = | τ | ∧ σ < L τ ) . That is, σ < LL τ if and only if σ is shorter than τ or σ and τ are the same length and
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Unformatted text preview: σ is smaller at the ﬁrst position they diﬀer. Compare this to the lexicographic ordering < L extended to all of 2 <ω , which is deﬁned by σ < L τ if and only if σ ⊂ τ or σ is less than τ at the ﬁrst position they diﬀer. For example 1 < LL 01 but 01 < L 1. It is easy to check that < LL well orders 2 <ω . However, < L is not a well order. For example, the set { n 1 | n ∈ N } ⊆ 2 <ω has no < L-least element. The deﬁnition of a ﬁnished tableau τ from a set of premises Σ requires that Tα must appear on every noncontradictory path through τ for every α in Σ. Technically you also check the CST has this property. The check is simple. Every noncontradictory path in τ extends some noncontradictory path in τ m , the tableau we had at stage m , and at stage m we added Tα m to the end of each such path when we form τ m +1 . 1...
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