Chapter_10 - PREDICATE LOGIC DERIVATIONS So far we have...

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PREDICATE LOGIC: DERIVATIONS So far, we have learned two systems for doing derivations in SL : SD and SD+ . We now need a derivation system for sentences of PL . Our first such system is called PD (‘predicate derivations’). PD = all the rules of SD (not SD+ ) plus an introduction and elimination rule for each quantifier. SD : R, ~I, ~E, &I, &E, I, E, I, E, I, E PD : all of these, plus 2200 I, 2200 E, 5 I, 5 E Two of these new rules are easy: Universal Elimination ( 2200 E) ( 2200 x) P P (a/x) If a sentence is true of everything in the UD, then it is true of all substitution instances. (Clearly, this is truth- preserving.)
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Example: All athletes are fit. Mary is an athlete. Mary is fit. 1. ( 2200 x)(Ax Fx) Assumption 2. Am Assumption 3. Am Fm 1 2200 E 4. Fm 2, 3 E There are no restrictions on your choice of constant. Examples: 1. ( 2200 x)Fxa Assumption 2. Fca 1 2200 E 1. ( 2200 x)Fxa Assumption 2. Faa 1 2200 E Assume: UD: positive integers Fxy: x is greater than or equal to y a: 1
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Existential Introduction ( 5 I) P (a/x) ( 5 x) P If a sentence is true of a certain member of the UD, then it is true of some member of the UD; it is true of something in the UD. (This too is clearly truth- preserving.) Jean Chretien is a politician Someone is a politician 1. Pc Assumption 2. ( 5 x)Px 1 5 I You don’t need to replace every occurrence of the constant being generalized: Harry is as tall as Harry. 1. Thh Assumption 2. ( 5 x)Txh 1 5 I 3. ( 5 x)Thx 1 5 I 4. ( 5 x)Txx 1 5 I
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But note: 1. Fac Assumption 2. ( 5 x)Fxx1 5 I MISTAKE! From the fact that Allan is Christine’s father it does not follow that something is its own father. So, each constant must be instantiated with a separate variable. 1. Fac Assumption 2. ( 5 x)( 5 y)Fxy 1 5 I
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Universal Introduction ( 2200 I) 5 I and 2200 E are easy rules to understand and to apply. 2200 I and 5 E, however, are more difficult. All bears are mammals All mammals have lungs All bears have lungs. Why does this argument seem so appealing? Well, consider the following informal reasoning. Take any arbitrary member of the UD – let’s call it b. If b is a bear then b is a mammal (by premise 1). If b is a mammal, then b has lungs (by premise 2). So, if b is a bear, then b has lungs. Clearly, this reasoning works for any arbitrary member of the UD. P (a/x) ( 2200 x) P Provided that: (1) ‘a’ does not occur in an undischarged assumption (2) ‘a’ does not occur in ( 2200 x) P .
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All bears are mammals. All mammals have lungs. All bears have lungs. 1. ( 2200 x)(Bx Mx) Assumption 2. ( 2200 x)(Mx Lx) Assumption 3. Ba Ma 1 2200 E 4. Ma La 1 2200 E 5. Ba Assumption 6. Ma 3, 5 E 7. La 4, 6 E 8. Ba La 5-7 I 9. ( 2200 x)(Bx Lx) 8 2200 I Both provisions are satisfied. The restrictions on
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This note was uploaded on 09/28/2009 for the course PHL 245 taught by Professor Bangu during the Winter '05 term at University of Toronto.

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Chapter_10 - PREDICATE LOGIC DERIVATIONS So far we have...

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