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Proofs about what can be proven implies that you can tell the limits of what you ... bird. Logic Notation. Bird (Albatross) Bird (Squigs) Flies(Squigs) Lays ...

Logic

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**Slide 1:**Logic

Mature, centuries of development, universally understood Proofs about what can be proven implies that you can tell the limits of what you can and cannot do But: Distracted by the mathematics of logic, instead of problem to be solved Many problems not amenable to logic techniques

**Slide 2:**Notation

If the animal has feathers Then the animal is a bird If the animal flies It lays eggs Then the animal is a bird

**Slide 3:**Logic Notation

Bird (Albatross) Bird (Squigs) Flies(Squigs) Lays-eggs(Squigs) Flies(Squigs) & Lays-eggs(Squigs)

**Slide 4:**Predicates

Functions that map object arguments to true or false Flies(Squigs) V Lays-eggs(Squigs) Squigs is an object that satisfies either or both predicates; that is Squigs is a symbol for which either or both predicates is true

**Slide 5:**Connectives

&, V, not, ? Feathers(Suzie) ? Bird(Suzie)

**Slide 6:**Truth tables

**Slide 7:**Properties

Commutative Associative Distributive DeMorgan’s laws !(!E) ?? E

**Slide 8:**Quantifiers

Universal A(x)[Feathers(x) ? Bird(x)] If the above expression is true it implies that you get a true expression when you substitute any object for x inside the square brackets Existential E(x) [Bird(x)] There exists at least one object substitutable for x inside the square brackets that makes Bird(x) true

**Slide 9:**Vocabulary of Logic

Objects + Variables == Terms Terms + Predicates == Atomic Formulas Atomic formulas + negation == Literals Literals + Connectives + quantifiers == wffs Well formed formulas (wffs) Sentences (all variables bound) A(x)[Feathers(x) V !Feathers(y)] Y is not bound

**Slide 10:**First order-predicate calculus

Variables can only represent objects Variables cannot represent predicates (2nd order predicate calculus) No variables (Propositional calculus) Clause: disjunction of literals

**Slide 11:**Interpretation

Objects in a world correspond to object symbols in logic Relations in a world correspond to predicates in logic Interpretation: Full accounting of the correspondence between objects and object symbols and between relations and predicates

**Slide 12:**Proofs

Feathers(squigs) A(x)[Feathers(x) ? Bird(x)] Axioms. Interpretation is a model for the above expressions. Show that all interpretations that make the axioms true also make Bird(squigs) true: you have been asked to prove that Bird(squigs) is a theorem wrt the axioms above Prove: Bird(squigs)

**Slide 13:**Proof procedures

Proof procedures use legal manipulations of symbols and predicates to produce new expressions from old expressions Sound rules of inference: legal manipulations Procedure: Apply sound rules of inference to axioms, and to the results of applying sound rules of inference, .. Until the desired theorem appears

**Slide 14:**Modus ponens

Sound rule of inference called Modus Ponens If there is an axiom of the form E1?E2, and there is another axiom of the form E1, then E2 logically follows If E2 is the theorem you want to prove, you are done, otherwise add E2 to the list of axioms, because E2 will always be true when all the rest of the axioms are true

**Slide 15:**Proof

A(x)[Feathers(x) ? Bird(x)] Feathers(squigs) Specialize: Feathers(squigs) ? Bird(squigs) E1 ??Feathers(squigs) already an axiom E2 ??Bird(squigs) logically follows using Modus Ponens (a sound rule of inference)

**Slide 16:**Proofs

Marcus is a man Marcus is a pompein Marcus was born in 40AD All men are mortal All pompeins died when the volcano erupted in 79AD No mortal lives longer than 150 years It is now 2004

**Slide 17:**Facts about Marcus

Man(marcus) Pompein(marcus) Born(marcus, 40) A(x) [man(x) ?mortal(x)] Erupted(Volcano, 79) A(x) [Pompein(x) ? died(x, 79)] A(x) A(t1) A(t2) [mortal(x) & born(x, t1) & gt(t2 – t1, 150) ? dead(x, t2)] Now = 2004

**Slide 18:**Prove marcus is dead

Killed by volcano More than 150 years Let’s try to word backwords

**Slide 19:**More marcus world facts

dead means not alive A(x) A(t) [dead(x, t) ? !alive(x, t)] Once you die, you stay dead A(x)A(t1) A(2)[died(x, t1) & gt(t2, t1) ? dead(x, t2)]

**Slide 20:**Facts about Marcus

Man(marcus) Pompein(marcus) Born(marcus, 40) A(x) [man(x) ?mortal(x)] Erupted(Volcano, 79) A(x) [Pompein(x) ? died(x, 79)] A(x) A(t1) A(t2) [mortal(x) & born(x, t1) & gt(t2 – t1, 150) ? dead(x, t2)] Now = 2004 A(x) A(t) [dead(x, t) ? !alive(x, t)] A(x)A(t1) A(2)[died(x, t1) & gt(t2, t1) ? dead(x, t2)]

**Slide 21:**Resolution

Resolution is a sound rule of inference If E1 V E2 !E2 V E3 Then E1 V E3 logically follows

**Slide 22:**Resolution Proof (1)

Feathers(Squigs) A(x) [Feathers(x) ? Bird(x)] Specialize Feathers(Squigs) Feathers(Squigs) ? Bird(Squigs) Rewrite Feathers(Squigs) !Feathers(Squigs) V Bird(Squigs) Resolve E1 V E2 !E2 V E3 E1 V E3 What are E1, E2, E3?

**Slide 23:**Resolution is a sound rule of inference

Subsumes Modus Ponens If there is an axiom of the form E1 ? E2, and Another axiom of the form E1 Then E2 logically follows Subsumes Modus Tolens E1 ? E2 !E2 Then !E1 logically follows

**Slide 24:**Resolution Proofs (2) Resolution Proof by Refutation

Assume that the negation of the theorem is T Show that the axioms and the assumed negation of the Theorem leads to a contradiction Conclude that the assumed negation of the theorem cannot be true because it leads to a contradiction Conclude that the Theorem must be true because the assumed negation of the theorem cannot be true

**Slide 25:**Refutation

A(x)[Feathers(x) ? Bird(x)] Feathers(squigs) Specialize: Feathers(squigs) ? Bird(squigs) Feathers(squigs)

**Slide 26:**Proof cont’d

Remove ? and rewrite !Feathers(squigs) V Bird(squigs) Feathers(squigs) Add negation of theorem to be proven !Bird(squigs) !Feathers(squigs) V Bird(squigs) Feathers(squigs)

**Slide 27:**Resolve 2 and 3

!Bird(squigs) !Feathers(squigs) V Bird(squigs) Feathers(squigs) RESOLVE !Bird(squigs) !Feathers(squigs) V Bird(squigs) Feathers(squigs) Bird(squigs) Contradiction !Bird(squigs) Bird(squigs) Contradiction! Therefore…Nil, Therefore !Bird(squigs) must be false, therefore Bird(squigs) must be true