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Metatheorems Computational Logic Lecture 8
Michael Genesereth Autumn 2011
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Formal Proofs A formal proof of from is a sequence of sentences terminating in in which each item is either: 1. a premise (a member of ) 2. an instance of an axiom schema 3. the result of applying a rule of inference to earlier items in the sequence.
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Metatheorems Deduction Theorem: |- ( ) if and only if {} |- . Substitution Theorem: |- ( ) and |- , then it is the case that |- . Chaining Theorem: If |- ( ) and |- ( ), then |- ( ).
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Proof Without Metatheorems
Problem: {p q, q r} |- (p r)?
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Proof Using Deduction Theorem
Problem: {p q, q r} |- (p r)?
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TA Appeasement Rules When we ask you to show that something is true, you may use metatheorems. When we ask you to give a formal proof, it means you should write out the proof as defined above. When we ask you to give a formal proof using certain rules of inference or axiom schemata, it means you should do so using only those rules of inference and axiom schemata and no others.
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Propositional Metatheorems
Propositional Deduction Theorem: |- ( ) if and only if {} |- . Propositional Substitution Theorem: |- ( ) and |- , then it is the case that |- . Propositional Chaining Theorem: If |- ( ) and |- ( ), then |- ( ).
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Results Bad News: As stated, none of these hold for Relational Logic.
Good News: Variations of these metatheorems do hold.
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Deduction Theorem Propositional Deduction Theorem: |- ( ) if and only if {} |- . : {} : p(x) : x.p(x) It is easy to show that {p(x)} |- x.p(x). One application of Universal Generalization. What about |- (p(x) x.p(x))? This is equivalent to |- (x.p(x) x.p(x))? Obviously, can be false.
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Relational Deduction Theorem
Relational Deduction Theorem: If has no free variables, then |- ( ) if and only if {} |- .
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