Metatheorems Computational Logic Lecture 8 Michael Genesereth Autumn 2011
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.
Metatheorems Deduction Theorem: |- ( ) if and only if {} |- . Substitution Theorem: |- ( ) and |- , then it is the case that |- . Chaining Theorem: If |- ( ) and |- ( ), then |- ( ).
Proof Without Metatheorems Problem: {p q, q r} |- (p r)?
Proof Using Deduction Theorem Problem: {p q, q r} |- (p r)?
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.
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 |- ( ).
Results Bad News: As stated, none of these hold for Relational Logic. Good News: Variations of these metatheorems do hold.
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.
Relational Deduction Theorem Relational Deduction Theorem: If has no free variables, then |- ( ) if and only if {} |- .