Positive Logic Is 2-Exptime Hard Aleksy Schubert Pawe Urzyczyn - PowerPoint PPT Presentation

Positive Logic Is 2-Exptime Hard Aleksy Schubert Paweł Urzyczyn Daria Walukiewicz-Chrząszcz University of Warsaw TYPES, Toulouse, April 26, 2013

Work in Progress. . .

Positive Logic Is 2-UExptime Hard Aleksy Schubert Paweł Urzyczyn Daria Walukiewicz-Chrząszcz University of Warsaw TYPES, Toulouse, April 26, 2013

Positive Logic Is 2-co-NExptime Hard Aleksy Schubert Paweł Urzyczyn Daria Walukiewicz-Chrząszcz University of Warsaw TYPES, Toulouse, April 26, 2013

Motivation Foundational research on: ◮ Expressive power of “weak” intuitionistic logics. ◮ High-level properties of proof tactics.

Prenex normal form In classical logic: Every formula is classically equivalent to one of the form: Q 1 x 1 Q 2 x 2 . . . Q k x k . Body ( x , x 2 , . . . , x k ) , where Body has no quantifiers.

Prenex normal form In classical logic: Every formula is classically equivalent to one of the form: Q 1 x 1 Q 2 x 2 . . . Q k x k . Body ( x , x 2 , . . . , x k ) , where Body has no quantifiers. Classification: Formulas may be classified according to the quantifier prefix, e.g. universal ( ∀ ∗ ) formulas are in Π 1 , and Π 2 is ∀ ∗ ∃ ∗ .

Prenex normal form In classical logic: Every formula is classically equivalent to one of the form: Q 1 x 1 Q 2 x 2 . . . Q k x k . Body ( x , x 2 , . . . , x k ) , where Body has no quantifiers. Classification: Formulas may be classified according to the quantifier prefix, e.g. universal ( ∀ ∗ ) formulas are in Π 1 , and Π 2 is ∀ ∗ ∃ ∗ . In intuitionistic logic:

Prenex normal form In classical logic: Every formula is classically equivalent to one of the form: Q 1 x 1 Q 2 x 2 . . . Q k x k . Body ( x , x 2 , . . . , x k ) , where Body has no quantifiers. Classification: Formulas may be classified according to the quantifier prefix, e.g. universal ( ∀ ∗ ) formulas are in Π 1 , and Π 2 is ∀ ∗ ∃ ∗ . In intuitionistic logic: The prenex fragment is decidable in Pspace.

The language we study To make things simpler, we consider first-order formulas ◮ with universal quantifiers and implications; ◮ without function symbols This fragment is known to be undecidable.

Mints Hierarchy Can we restore the prenex classification in intuitionistic logic?

Mints Hierarchy Can we restore the prenex classification in intuitionistic logic? Grigori Minc (1968): Yes, consider the quantifier prefix a formula would get if classically normalized.

Mints Hierarchy Can we restore the prenex classification in intuitionistic logic? Grigori Minc (1968): Yes, consider the quantifier prefix a formula would get if classically normalized. For example, ∀ quantifiers occurring at positive positions will remain ∀ in the prefix.

Positive and Negative + – + + – – + – + + – + – – +

Mints Hierarchy Π 1 – All quantifiers at positive positions. Σ 1 – All quantifiers at negative positions. Π 2 – One alternation: some negative quantifiers in scope of some positive ones. Σ 2 – One alternation: some positive quantifiers in scope of some negative ones. And so on.

Lower bounds for Mints Hierarchy Π 1 – 2-UExptime -hard Σ 1 – At least Exptime -hard Π 2 – Undecidable Σ 2 – Undecidable Work in progress: with function symbols ◮ Class Σ 1 becomes undecidable. ◮ Class Π 1 is of the same complexity as before.

A positive example Let ϕ = ( I → Tv ) → Ap ( x ) , ψ = I → ( D → Tv ) → Ap ( x ) and ϑ = D → Ap ( x ) , and prove the formula ( ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ) → Tv .

A positive example Let ϕ = ( I → Tv ) → Ap ( x ) , ψ = I → ( D → Tv ) → Ap ( x ) and ϑ = D → Ap ( x ) , and prove the formula ( ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ) → Tv . The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv

A positive example Let ϕ = ( I → Tv ) → Ap ( x ) , ψ = I → ( D → Tv ) → Ap ( x ) and ϑ = D → Ap ( x ) , and prove the formula ( ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ) → Tv . The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x ))

A positive example Let ϕ = ( I → Tv ) → Ap ( x ) , ψ = I → ( D → Tv ) → Ap ( x ) and ϑ = D → Ap ( x ) , and prove the formula ( ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ) → Tv . The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x )) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ϕ → ψ → ϑ → Ap ( x )

A positive example Let ϕ = ( I → Tv ) → Ap ( x ) , ψ = I → ( D → Tv ) → Ap ( x ) and ϑ = D → Ap ( x ) , and prove the formula ( ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ) → Tv . The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x )) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ϕ → ψ → ϑ → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Ap ( x ) ϕ, ψ, ϑ

Example continued The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Ap ( x ) ϕ = ( I → Tv ) → Ap ( x ) , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x )

Example continued The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Ap ( x ) ϕ = ( I → Tv ) → Ap ( x ) , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x )

Example continued The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Ap ( x ) ϕ = ( I → Tv ) → Ap ( x ) , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv I → Tv ϕ = ( I → Tv ) → Ap ( x ) , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x )

Example continued The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Ap ( x ) ϕ = ( I → Tv ) → Ap ( x ) , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv I → Tv ϕ = ( I → Tv ) → Ap ( x ) , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x ) I , ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ϕ = ( I → Tv ) → Ap ( x ) , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x )

Example continued The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ϕ = ( I → Tv ) → Ap ( x ) I , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x )

Example continued The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ϕ = ( I → Tv ) → Ap ( x ) I , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x )) ϕ ( x ) = ( I → Tv ) → Ap ( x ) I , ϑ ( x ) = D → Ap ( x ) ψ ( x ) = I → ( D → Tv ) → Ap ( x )

Example continued The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ϕ = ( I → Tv ) → Ap ( x ) I , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x )) ϕ ( x ) = ( I → Tv ) → Ap ( x ) I , ϑ ( x ) = D → Ap ( x ) ψ ( x ) = I → ( D → Tv ) → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ϕ ( x ′ ) → ψ ( x ′ ) → ϑ ( x ′ ) → Ap ( x ′ ) I , ϕ ( x ) , ϑ ( x ) , ψ ( x )

Example continued The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ϕ = ( I → Tv ) → Ap ( x ) I , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x )) ϕ ( x ) = ( I → Tv ) → Ap ( x ) I , ϑ ( x ) = D → Ap ( x ) ψ ( x ) = I → ( D → Tv ) → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ϕ ( x ′ ) → ψ ( x ′ ) → ϑ ( x ′ ) → Ap ( x ′ ) I , ϕ ( x ) , ϑ ( x ) , ψ ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Ap ( x ′ ) I , ϕ ( x ) , ϑ ( x ) , ψ ( x ) ϕ ( x ′ ) , ϑ ( x ′ ) , ψ ( x ′ )

Example continued The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ϕ = ( I → Tv ) → Ap ( x ) I , ϑ = D → Ap ( x ) ψ = I → ( D → Tv ) → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ∀ x ( ϕ → ψ → ϑ → Ap ( x )) ϕ ( x ) = ( I → Tv ) → Ap ( x ) I , ϑ ( x ) = D → Ap ( x ) ψ ( x ) = I → ( D → Tv ) → Ap ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv ϕ ( x ′ ) → ψ ( x ′ ) → ϑ ( x ′ ) → Ap ( x ′ ) I , ϕ ( x ) , ϑ ( x ) , ψ ( x ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Ap ( x ′ ) I , ϕ ( x ) , ϑ ( x ) , ψ ( x ) ϕ ( x ′ ) , ϑ ( x ′ ) , ψ ( x ′ )

The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Ap ( x ′ ) I , ϕ ( x ) , ϑ ( x ) , ψ ( x ) ϕ ( x ′ ) = ( I → Tv ) → Ap ( x ′ ) ϑ ( x ′ ) = D → Ap ( x ′ ) ψ ( x ′ ) = I → ( D → Tv ) → Ap ( x ′ )

The Assumptions The Goal ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Ap ( x ′ ) I , ϕ ( x ) , ϑ ( x ) , ψ ( x ) ϕ ( x ′ ) = ( I → Tv ) → Ap ( x ′ ) ϑ ( x ′ ) = D → Ap ( x ′ ) ψ ( x ′ ) = I → ( D → Tv ) → Ap ( x ′ ) ∀ x ( ϕ → ψ → ϑ → Ap ( x )) → Tv Tv ϕ ( x ) , ϑ ( x ) , ψ ( x ) I , D , ϕ ( x ′ ) = ( I → Tv ) → Ap ( x ′ ) ϑ ( x ′ ) = D → Ap ( x ′ ) ψ ( x ′ ) = I → ( D → Tv ) → Ap ( x ′ )