introduction to logic herbrand logic proofsintrologic.stanford.edu/lectures/lecture_12.pdfuniversal...

Introduction to LogicHerbrand Logic Proofs

Michael GeneserethComputer Science Department

Stanford University

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Propositional Logic

{m ⇒ p ∨ q, p ⇒ q} ⊨ m ⇒ q?

Relational Logic

{p(a) ∨ p(b), ∀x.(p(x) ⇒ q(x))} ⊨ ∃x.q(x)?

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Object constants: nBinary relation constants: kFactoids in Herbrand Base: infiniteInterpretations: infinite

Herbrand Logic

Herbrand Proofs

Logical Rules of Inference

Negation IntroductionNegation Elimination

And IntroductionAnd Elimination

Or IntroductionOr Elimination

AssumptionImplication Elimination

Biconditional IntroductionBiconditional Elimination

Universal IntroductionUniversal Elimination

Existential IntroductionExistential Elimination

Domain ClosureInduction (Linear, Tree, Structural) New!!

Rules of Inference for Quantifiers

A structured proof of a conclusion from a set of premises is a sequence of (possibly nested) sentences terminating in an occurrence of the conclusion at the top level of the proof. Each step in the proof must be either (1) a premise (at the top level), (2) an assumption, or (3) the result of applying an ordinary or structured rule of inference to earlier items in the sequence (subject to the constraints given above).

If there exists a proof of a sentence φ from a set Δ of premises using the rules of inference in Herbrand, we say that φ is provable from Δ (written Δ ⊢Herbrand φ).

Structured Proof

Example - Spin

On transition, a particle goes from spin zero to spin positive or spin negative and then goes to spin zero on second transition. Show that if it has spin zero, it will have spin zero after two transitions.

Givens: ∀x:(zero(x) ⇒ pos(f(x)) | neg(f(x))) ∀x:(pos(x) ⇒ zero(f(x))) ∀x:(neg(x) ⇒ zero(f(x)))

Goal: ∀x:(zero(x) ⇒ zero(f(f(x))))

Problem

1. ∀x.(zero(x) ⇒ pos(f(x)) ∨ neg(f(x))) Premise2. ∀x.(pos(x) ⇒ zero(f(x))) Premise3. ∀x.(neg(x) ⇒ zero(f(x))) Premise

Proof

1. ∀x.(zero(x) ⇒ pos(f(x)) ∨ neg(f(x))) Premise2. ∀x.(pos(x) ⇒ zero(f(x))) Premise3. ∀x.(neg(x) ⇒ zero(f(x))) Premise4. zero(x) Assumption

Proof

1. ∀x.(zero(x) ⇒ pos(f(x)) ∨ neg(f(x))) Premise2. ∀x.(pos(x) ⇒ zero(f(x))) Premise3. ∀x.(neg(x) ⇒ zero(f(x))) Premise4. zero(x) Assumption5. zero(x) ⇒ pos(f(x)) ∨ neg(f(x)) UE: 1

Proof

1. ∀x.(zero(x) ⇒ pos(f(x)) ∨ neg(f(x))) Premise2. ∀x.(pos(x) ⇒ zero(f(x))) Premise3. ∀x.(neg(x) ⇒ zero(f(x))) Premise4. zero(x) Assumption5. zero(x) ⇒ pos(f(x)) ∨ neg(f(x)) UE: 16. pos(f(x)) ∨ neg(f(x)) IE: 5, 4

Proof

1. ∀x.(zero(x) ⇒ pos(f(x)) ∨ neg(f(x))) Premise2. ∀x.(pos(x) ⇒ zero(f(x))) Premise3. ∀x.(neg(x) ⇒ zero(f(x))) Premise4. zero(x) Assumption5. zero(x) ⇒ pos(f(x)) ∨ neg(f(x)) UE: 16. pos(f(x)) ∨ neg(f(x)) IE: 5, 47. pos(f(x)) ⇒ zero(f(f(x))) UE: 28. neg(f(x)) ⇒ zero(f(f(x))) UE: 3

Proof

1. ∀x.(zero(x) ⇒ pos(f(x)) ∨ neg(f(x))) Premise2. ∀x.(pos(x) ⇒ zero(f(x))) Premise3. ∀x.(neg(x) ⇒ zero(f(x))) Premise4. zero(x) Assumption5. zero(x) ⇒ pos(f(x)) ∨ neg(f(x)) UE: 16. pos(f(x)) ∨ neg(f(x)) IE: 5, 47. pos(f(x)) ⇒ zero(f(f(x))) UE: 28. neg(f(x)) ⇒ zero(f(f(x))) UE: 39. zero(f(f(x))) OE: 6, 7, 8

Proof

1. ∀x.(zero(x) ⇒ pos(f(x)) ∨ neg(f(x))) Premise2. ∀x.(pos(x) ⇒ zero(f(x))) Premise3. ∀x.(neg(x) ⇒ zero(f(x))) Premise4. zero(x) Assumption5. zero(x) ⇒ pos(f(x)) ∨ neg(f(x)) UE: 16. pos(f(x)) ∨ neg(f(x)) IE: 5, 47. pos(f(x)) ⇒ zero(f(f(x))) UE: 28. neg(f(x)) ⇒ zero(f(f(x))) UE: 39. zero(f(f(x))) OE: 6, 7, 8

10. zero(x) ⇒ zero(f(f(x)))) II: 4, 9

Proof

1. ∀x.(zero(x) ⇒ pos(f(x)) ∨ neg(f(x))) Premise2. ∀x.(pos(x) ⇒ zero(f(x))) Premise3. ∀x.(neg(x) ⇒ zero(f(x))) Premise4. zero(x) Assumption5. zero(x) ⇒ pos(f(x)) ∨ neg(f(x)) UE: 16. pos(f(x)) ∨ neg(f(x)) IE: 5, 47. pos(f(x)) ⇒ zero(f(f(x))) UE: 28. neg(f(x)) ⇒ zero(f(f(x))) UE: 39. zero(f(f(x))) OE: 6, 7, 8

10. zero(x) ⇒ zero(f(f(x)))) II: 4, 911. ∀x.(zero(x) ⇒ zero(f(f(x)))) UI: 10

Proof

Analysis

A set of premises Δ logically entails a conclusion ϕ (Δ |= ϕ) if and only if every interpretation that satisfies Δ also satisfies ϕ.

If there exists a proof of a sentence φ from a set Δ of premises using the rules of inference in R, we say that φ is provable from Δ using R (written Δ ⊢R φ).

Logical Entailment and Provability

A proof system is sound if and only if every provable conclusion is logically entailed.

If Δ ⊢ φ, then Δ ⊨ φ.

A proof system is complete if and only if every logical conclusion is provable.

If Δ ⊨ φ, then Δ ⊢ φ.

Soundness and Completeness

Theorem: Fitch is sound and complete for Relational Logic.

Δ |= ϕ if and only if Δ ⊢Fitch φ.

Theorem: Herbrand is sound for Herbrand Logic.

if Δ ⊢Herbrand φ, then Δ |= ϕ.

Theorem: Fitch is not complete for Herbrand Logic.

However, everything that can be proved by Fitch can be proved by Herbrand!

Herbrand Proof System

HL is highly expressive.

For example, possible to axiomatize arithmetic as a finite set of axioms.

This is not possible with Relational Logic and in other logics (e.g. First-Order Logic).

Good News

The question of unsatisfiability and the question of logical entailment for HL are not even semidecidable.

Bad News

Diophantine equation:

3x2 = 1

Diophantine equation in Peano Arithmetic:

∃x.∃y.(times(x,x,y) ∧ times(s(s(s(0))),y,s(0)))

Unsolvability Question:

Axioms of Arithmetic ∪ {∃x.∃y.(times(x,x,y) ∧ times(s(s(s(0))),y,s(0)))}

Diophantine Equations

It is known that the question of unsolvability of Diophantine equations is not semidecidable.

Since we can represent the question of unsolvability of Diophantine equations in Herbrand Logic, if we could determine the unsatisfiability of sentences in Herbrand Logic, we could answer questions of unsolvability of Diophantine equations.

HL Not Even Semidecidable